Extra Credit 26

Q17.3.5

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

Step 1: Determine the equations from the information above. Cadmium metal corresponds to \(Cd(s)\) and it is oxidized, which means it loses electrons, to 1 M cadmium II, which is defined as \(Cd^{2+}(aq)\). This is the oxidation reaction so it occurs as the anode. The 1 M aluminum nitrate solution corresponds to \(Al(NO_{3})_{3}\) but since Al is the reduction element, only Al needs to be in the final reaction. If you remove the \((NO_{3})_{3}\) the Al will now have a 3+ charge. \(Al^{3+}(aq)\) will be reduced, which means it gains electrons, to \(Al(s)\). With this information we can write the final equation:

\[Cd(s)+Al^{3+}(aq)\rightarrow Cd^{2+}(aq)+Al(s)\]

Step 2: Separate the half reactions into their corresponding redox reactions. The oxidation reaction is the one containing cadmium because Cd goes from 0 charge to a 2+ charge. The reduction reaction is the one containing aluminum because Al goes from a 3+ charge to 0 charge.

Oxidation Equation: \[Cd(s)\rightarrow Cd^{2+}(aq)\]

Reduction Equation: \[Al^{3+}(aq)\rightarrow Al(s)\]

Step 3: These reactions only need to be balanced by electrons since they contain no O or H. The oxidation reaction has 0 charge on the left side and a 2+ charge on the right side. To balance the reaction we need to add 2 electrons to the right side.

Oxidation Reaction: \[Cd(s)\rightarrow Cd^{2+}(aq)+2e^{-}\]

The reduction reaction has 3+ charge on the left side and a 0 charge on the right side. To balance the equation we need to add 3 electrons to the left side.

Reduction Reaction: \[3e^{-}+Al^{3+}(aq)\rightarrow Al(s)\]

Step 4: Scale the reactions so that the electrons are equal. The oxidation reaction has 2 electrons and the reduction reaction has 3 electrons. To scale, we need to multiply the oxidation reaction by 3 and the reduction reaction by 2.

Oxidation Reaction: \[3[Cd(s)\rightarrow Cd^{2+}(aq)+2e^{-}]\Rightarrow 3Cd(s)\rightarrow 3Cd^{2+}(aq)+6e^{-}\]

Reduction Reaction: \[2[3e^{-}+Al^{3+}(aq)\rightarrow Al(s)]\Rightarrow 6e^{-}+2Al^{3+}(aq)\rightarrow 2Al(s)\]

Step 5: Add the reactions and cancel out common terms.

\[6e^{-}+2Al^{3+}(aq)+3Cd(s)\rightarrow 2Al(s)+3Cd^{2+}(aq)+6e^{-}\]

The electrons cancel out to finalize the complete balanced chemical equation of the original redox reaction

\[2Al^{3+}(aq)+3Cd(s)\rightarrow 2Al(s)+3Cd^{2+}(aq)\]

Step 6: Now that you have the balanced equation, use the standard reduction potentials for the oxidation reaction, -0.403, and reduction reaction, -1.66, to solve for \(E^{o}_{cell}\) using this equation:

\[E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}\]

To find the reduction potentials for each half reaction look on the list of reduction potentials that is often added to tests or in the back of textbooks, like this one. Always make sure that the table is at the right temperature at that the table lists reduction not oxidation. Note that metals that are more likely to be reduced are listed higher on the table, meaning that they are more likely to gain an electron. The farther two metals are away from each other on the table can also give an indication of the likelihood for spontaneity.

The cathode corresponds to the reduction reaction cell potential and the anode corresponds to the oxidation reaction cell potential. With this knowledge plug in the information to find E^ocell:

\[E^{o}_{cell}=-1.66-(-0.403)\]
\[E^{o}_{cell}=-1.26\]

\(E^{o}_{cell}\)=-1.26V

Step 7: The final step is to determine the spontaneity of the reaction. Since \(E^{o}_{cell}\) is negative, \(\Delta G\) is positive, which corresponds to a nonspontaneous reaction.

You can prove that this is true by plugging your value for Standard Potential into the Equation:

\(\Delta G^{o} = -nFE^{o}_{cell}\)

\(\Delta G^{o} = -(6 mol)(96.485 \frac{KJ}{V mol})(-1.26V)\)

\(\Delta G^{o} = 729.4266 KJ > 0\)

Q19.1.24

Predict which will be more stable, [CrO₄]^2- or [WO₄]^2-and explain. Write a brief description of the following

[CrO₄]^2- is more stable. To get this answer you have to look at the which orbital each element is in. Chromium is in the 3d orbital and Tungsten is in the 5d orbital. The higher the energy level the more stable the element. For this reason we see that [CrO₄]^2- is more stable since it is in a lower energy level, 3d, than that of Tungsten, 5d.

Another key factor is the reactivity of the metal due to its number of valence electrons. However, both ions are polyatomic and composed of transition metals, which makes it kind of difficult to decipher. Usually polyatomic ions form because they are some of the most stable forms that specific metal can take.

Q12.4.17

Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for 1/64 of the initial dose to remain in the athlete’s body?

Step 1: From the problem we are given that this process is first order. To find this specific time we have to find the relative k value by the first order half life equation. We are given the half life as 42 days, and that is essentially the only information we need for the formula. When we solve for k we get:

Half life first order reaction: \[t_{1/2}=\frac{ln2}{k}\]
Plug in half life: \[42=\frac{0.693}{k}\]
Solve for k: \[k=\frac{0.693}{42}\]
\[k=0.0165\]

Step 2: Now that we know k=0.0165 we can use this value along with the desired dose amount, 1/64, in the first order reaction equation to solve for the time it would take for the athlete to reach this dose. Solving for this time we get:

First order reaction: \[[A]=[A]_{0}e^{-kt}\]
Plug in the desired dose amount (in other words, solve figure out the proportion of [A]₀ that is equivalent to [A]: \[[A]=\frac{1}{64}[A]_{0}\]

You could also think of this differently mathematically. 1/64 is equal to .015625 or about 1.5625% of the original dose, which would make [A]₀ = 100% and [A] = 1.5625%.
Substitute for [A]: \[\frac{1}{64}[A]_{0}=[A]_{0}e^{-0.0165t}\]

[A]₀ cancels out and: \[\frac{1}{64}=e^{-0.0165t}\]
Take the natural log of both sides: \[ln(\frac{1}{64})=-0.0165t\]
Solve for t: \[-4.159=-0.0165t\]
\[t=\frac{-4.159}{-0.0165}\]
\[t=252 days\]

If you used the percentage method, you could use the equation: \[ln(\frac{A_{t}}{A_{0}}) = -kt\]

Then plug in you values and solve for t. It should give you the same answer.

\[ln(\frac{1.5625%}{100%}) = -(0.0165)t\]

\[t = \frac{-ln(0.015625)}{0.0165}\]

\[t = 252.05 days\]

Answer: It would take 252 days for the athlete to have 1/64 of the initial dose remaining in their body.

Q21.3.1

Write a brief description of the following:

a) Nucleon: Any particle contained in the nucleus of the atom, which can refer to protons and neutrons.

b) Alpha particle: One product of natural radioactivity and is the nucleus of a helium atom.

c) Beta particle: A high-speed electron that is one product of natural radioactivity.

d) Positron: A positron is a particle with the same mass as an electron but with a positive charge.

e) Gamma rays: Compose electromagnetic radiation of high energy and short wavelength.

f) Nuclide: A term used when referring to a single type of nucleus.

g) Mass number: The sum of the number of protons and neutrons in an element.

h) Atomic Number: The nuber of protons in the nucleus of an element.

Q21.7.3

Given specimens uranium-232 (t_1/2=68.9y) and uranium-233 (t_1/2=159,200y) of equal mass, which one would have greater activity and why?

If you look at the equation defining activity, \(A=\lambda N\) , where \(\lambda\) is the decay constant given as the probability of decay per unit time and N being the number of parent nuclei. Since the number nuclei is very similar in both specimens we have to look at the half lives of each. The equation \(t_{1/2}=\frac{ln2}{\lambda }\) gives us corresponding \(\lambda\) for each specimen. Solving for each λ we get:

U-232: \[t_{1/2}=\frac{ln2}{\lambda }\]

\[{\lambda }=\frac{ln2}{t_{1/2}}\]

\[{\lambda }=\frac{ln2}{68.9}\]

\[{\lambda }=0.0101\]

U-233: \[t_{1/2}=\frac{ln2}{\lambda }\]

\[{\lambda }=\frac{ln2}{t_{1/2}}\]

\[{\lambda }=\frac{ln2}{159,200}\]

\[{\lambda }=4.35x10^{-6}\]

We use the equation \(A=\lambda N\) ,with no effect of N because they are the same mass, so \(A=\lambda\) to determine the species with the higher activity. Whichever specimen has the bigger lambda has a bigger activity and in this case it is U-232. Uranium-232 would have a greater activity.

Q20.4.13

Balance each reaction and calculate the standard electrode potential for each. Be sure to include the physical state of each product and reactant.

a) \(Cl_{2}(g)+H_{2}(g)\rightarrow 2Cl^{-}(aq)+2H^{+}(aq)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing hydrogen because H goes from 0 charge to a 2+ charge. The reduction reaction is the one containing chlorine because Cl goes from 0 charge to -2 charge.

Oxidation Equation: \[H_{2}(g)\rightarrow 2H^{+}(aq)\]

Reduction Equation: \[Cl_{2}(g)\rightarrow 2Cl^{-}(aq)\]

Step 2: These reactions only need to be balanced by electrons since they contain no O or H. The oxidation reaction has 0 charge on the left side and a 2+ charge on the right side. To balance the reaction we need to add 2 electrons to the right side.

Oxidation Reaction: \[H_{2}(g)\rightarrow 2H^{+}(aq)+2e^{-}\]

The reduction reaction has 0 charge on the left side and a 2- charge on the right side. To balance the equation we need to add 2 electrons to the left side.

Reduction Reaction: \[2e^{-}+Cl_{2}(g)\rightarrow 2Cl^{-}(aq)\]

Step 3: Scale the reactions so that the electrons are equal. In this problem, the electrons are already equal, so the reactions do not need to be scaled.

Step 4: Add the reactions and cancel out common terms.

\[2e^{-}+Cl_{2}(g)+H_{2}(g)\rightarrow 2Cl^{-}(aq)+2H^{+}(aq)+2e^{-}\]

The electrons cancel out to finalize the complete balanced chemical equation of the original redox reaction

\[Cl_{2}(g)+H_{2}(g)\rightarrow 2Cl^{-}(aq)+2H^{+}(aq)\]

Step 5: Now that you have the balanced equation, use the standard reduction potentials for the oxidation reaction, 0.000, and reduction reaction, 1.358, to solve for \(E^{o}_{cell}\) using this equation:

\[E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}\]

The cathode corresponds to the reduction reaction cell potential and the anode corresponds to the oxidation reaction cell potential. With this knowledge plug in the information to find \(E^{o}_{cell}\):

\[E^{o}_{cell}=1.358-(0.000)\]
\[E^{o}_{cell}=1.358\]

\(E^{o}_{cell}\)=1.358 V

b) \(Br_{2}(aq)+Fe^{2+}(aq)\rightarrow 2Br^{-}(aq)+Fe^{3+}(aq)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing iron because Fe goes from 2+ charge to a 3+ charge. The reduction reaction is the one containing bromine because Br goes from 0 charge to 2- charge.

Oxidation Equation: \[Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)\]

Reduction Equation: \[Br_{2}(aq)\rightarrow 2Br^{-}(aq)\]

Step 2: These reactions only need to be balanced by electrons since they contain no O or H. The oxidation reaction has 2+ charge on the left side and a 3+ charge on the right side. To balance the reaction we need to add 1 electrons to the right side.

Oxidation Reaction: \[Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+e^{-}\]

The reduction reaction has 0 charge on the left side and a 2- charge on the right side. To balance the equation we need to add 2 electrons to the left side.

Reduction Reaction: \[2e^{-}+Br_{2}(aq)\rightarrow 2Br^{-}(aq)\]

Step 3: Scale the reactions so that the electrons are equal. The oxidation reaction has 1 electron and the reduction reaction has 2 electrons. To scale, we need to multiply the oxidation reaction by 2 and the reduction reaction by 1.

Oxidation Reaction: \[2[Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+e^{-}]\Rightarrow 2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2e^{-}\]

Reduction Reaction: \[1[2e^{-}+Br_{2}(aq)\rightarrow 2Br^{-}(aq)]\Rightarrow 2e^{-}+Br_{2}(aq)\rightarrow 2Br^{-}(aq)\]

Step 4: Add the reactions and cancel out common terms.

\[2e^{-}+Br_{2}(aq)+2Fe^{2+}(aq)\rightarrow 2Br^{-}(aq)+2Fe^{3+}(aq)+2e^{-}\]

The electrons cancel out to finalize the complete balanced chemical equation of the original redox reaction

\[Br_{2}(aq)+2Fe^{2+}(aq)\rightarrow 2Br^{-}(aq)+2Fe^{3+}(aq).\]

Step 5: Now that you have the balanced equation, use the standard reduction potentials for the oxidation reaction, 0.771, and reduction reaction, 1.0873, to solve for \(E^{o}_{cell}\) using this equation:

\[E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}\]

The cathode corresponds to the reduction reaction cell potential and the anode corresponds to the oxidation reaction cell potential. With this knowledge plug in the information to find \(E^{o}_{cell}\):

\[E^{o}_{cell}=1.087-(0.771)\]
\[E^{o}_{cell}=0.316\]

\(E^{o}_{cell}\)= 0.316 V

c) \(Fe^{3+}(aq)+Cd(s)\rightarrow Fe^{2+}(aq)+Cd^{2+}(aq)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing cadmium because Cd goes from 0 charge to a 2+ charge. The reduction reaction is the one containing iron because Fe goes from 3+ charge to 2+ charge.

Oxidation Equation: \[Cd(s)\rightarrow Cd^{2+}(aq)\]

Reduction Equation: \[Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)\]

Step 2: These reactions only need to be balanced by electrons since they contain no O or H. The oxidation reaction has 0 charge on the left side and a 2+ charge on the right side. To balance the reaction we need to add 2 electrons to the right side.

Oxidation Reaction: \[Cd(s)\rightarrow Cd^{2+}(aq)+2e^{-}\]

The reduction reaction has 3+ charge on the left side and a 2+ charge on the right side. To balance the equation we need to add 1 electron to the left side.

Reduction Reaction: \[e^{-}+Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)\]

Step 3: Scale the reactions so that the electrons are equal. The oxidation reaction has 2 electrons and the reduction reaction has 1 electron. To scale, we need to multiply the oxidation reaction by 1 and the reduction reaction by 2.

Oxidation Reaction: \[1[Cd(s)\rightarrow Cd^{2+}(aq)+2e^{-}]\Rightarrow Cd(s)\rightarrow Cd^{2+}(aq)+2Fe^{3+}(aq)+Cd(s)\rightarrow 2Fe^{2+}(aq)+Cd^{2+}(aq)\]

Reduction Reaction: \[2[e^{-}+Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)]\Rightarrow 2e^{-}+2Fe^{3+}(aq)\rightarrow 2Fe^{2+}(aq)\]

Step 4: Add the reactions and cancel out common terms.

\[2e^{-}+2Fe^{3+}(aq)+Cd(s)\rightarrow 2Fe^{2+}(aq)+Cd^{2+}(aq)+2e^{-}\]

The electrons cancel out to finalize the complete balanced chemical equation of the original redox reaction.

\[2Fe^{3+}(aq)+Cd(s)\rightarrow 2Fe^{2+}(aq)+Cd^{2+}(aq)\]

Step 5: Now that you have the balanced equation, use the standard reduction potentials chart to find the E^o for the oxidation reaction, -0.403, and reduction reaction, 0.771, to solve for E^o_cell using this equation:

\[E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}\]

The cathode corresponds to the reduction reaction cell potential and the anode corresponds to the oxidation reaction cell potential. With this knowledge plug in the information to find \(E^{o}_{cell}\):

\[E^{o}_{cell}=0.771-(-0.403)\]
\[E^{o}_{cell}=1.174\]

\(E^{o}_{cell}\)=1.174 V

Q20.4.15

Write a balanced chemical equation for each redox reaction:

A) \(H_{2}PO_{2}^{-}(aq) + SbO_{2}^{-}(aq) \rightarrow HPO_{3}^{2-}(aq) + Sb(s)\) in basic solution (fixed Mathjax)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing phosphorus because P goes from a 1+ charge to a 3+ charge. The reduction reaction is the one containing antimony because Sb goes from a 3+ charge to a 0 charge.

Oxidation Equation: \[H_{2}PO_{2}^{_{-}}(aq)\rightarrow HPO_{3}^{_{2-}}(aq)\]

Reduction Equation: \[SbO_{2}^{_{-}}(aq)\rightarrow Sb(s)\]

Step 2: Balance elements other than H and O. In this problem, everything but H and O is already balanced so we can move on to the next step.

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 1 water molecule while the reduction reaction needs to be balanced by adding 2 water molecules.

Oxidation Equation: \[H_{2}O(l)+H_{2}PO_{2}^{_{-}}(aq)\rightarrow HPO_{3}^{_{2-}}(aq)\]

Reduction Equation: \[SbO_{2}^{_{-}}(aq)\rightarrow Sb(s)+2H_{2}O(l)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 3 protons while the reduction reaction needs to be balanced by adding 4 protons.

Oxidation Equation: \[H_{2}O(l)+H_{2}PO_{2}^{_{-}}(aq)\rightarrow HPO_{3}^{_{2-}}(aq)+3H^{+}\]

Reduction Equation: \[4H^{+}+SbO_{2}^{_{-}}(aq)\rightarrow Sb(s)+2H_{2}O(l)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 1- charge on the left side and a (2-) + (3+) = 1+ charge on the right side. To balance the reaction we need to add 2 electrons to the right side.

Oxidation Reaction: \[H_{2}O(l)+H_{2}PO_{2}^{_{-}}(aq)\rightarrow HPO_{3}^{_{2-}}(aq)+3H^{+}+2e^{-}\]

The reduction reaction has a (4+) + (1-) = 3+ charge on the left side and no charge on the right side. To balance the equation we need to add 3 electrons to the left side.

Reduction Reaction: \[3e^{-}+4H^{+}+SbO_{2}^{_{-}}(aq)\rightarrow Sb(s)+2H_{2}O(l)\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 2 electrons and the reduction reaction has 3 electrons. Thus the oxidation reaction must be multiplied by 3 and the reduction reaction by 2.

Oxidation Reaction: \[3[H_{2}O(l)+H_{2}PO_{2}^{_{-}}(aq)\rightarrow HPO_{3}^{_{2-}}(aq)+3H^{+}+2e^{-}]\Rightarrow 3H_{2}O(l)+3H_{2}PO_{2}^{_{-}}(aq)\rightarrow 3HPO_{3}^{_{2-}}(aq)+9H^{+}+6e^{-}\]

Reduction Reaction: \[2[3e^{-}+4H^{+}+SbO_{2}^{_{-}}(aq)\rightarrow Sb(s)+2H_{2}O(l)]\Rightarrow 6e^{-}+8H^{+}+2SbO_{2}^{_{-}}(aq)\rightarrow 2Sb(s)+4H_{2}O(l)\]

Step 7: Add the reactions and cancel out common terms.

\[6e^{-}+8H^{+}+2SbO_{2}^{_{-}}(aq)+3H_{2}O(l)+3H_{2}PO_{2}^{_{-}}(aq)\rightarrow 2Sb(s)+4H_{2}O(l)+3HPO_{3}^{_{2-}}(aq)+9H^{+}+6e^{-}\]

The electrons cancel out as well as 8 protons and 3 water molecules

\[2SbO_{2}^{_{-}}(aq)+3H_{2}PO_{2}^{_{-}}(aq)\rightarrow 2Sb(s)+H_{2}O(l)+3HPO_{3}^{_{2-}}(aq)+H^{+}\]

Step 8: Since this reaction takes place in basic solutions, the final answer must contain hydroxide. For this step we can add on hydroxide to both sides of the equation to balance the protons. There is 1 net proton in this equation so we add 1 hydroxide to both sides.

\[OH^{-}+2SbO_{2}^{_{-}}(aq)+3H_{2}PO_{2}^{_{-}}(aq)\rightarrow 2Sb(s)+H_{2}O(l)+3HPO_{3}^{_{2-}}(aq)+H^{+}+OH^{-}\]

Step 9: Combine the hydroxide and proton ions present on the same side to form water.

\[OH^{-}+2SbO_{2}^{_{-}}(aq)+3H_{2}PO_{2}^{_{-}}(aq)\rightarrow 2Sb(s)+H_{2}O(l)+3HPO_{3}^{_{2-}}(aq)+H_{2}O(l)\]

Step 10: Cancel common terms to finalize the complete balanced chemical equation of the original redox reaction

\(OH^{-}+2SbO_{2}^{-}(aq)+3H_{2}PO_{2}^{-}(aq)\rightarrow 2Sb(s)+2H_{2}O(l)+3HPO_{3}^{2-}(aq)\)

B) \(HNO_{2}(aq) + I^{−}(aq) \rightarrow NO(g) + I^{2}(s)\) in acidic solution (fixed Mathjax)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing nitrogen because N goes from a 3+ charge to a 2+ charge. The reduction reaction is the one containing iodine because I goes from a 1- charge to a 0 charge.

Oxidation Reaction: \[HNO_{2}(aq)\rightarrow NO(g)\]

Reduction Reaction: \[I^{-}(aq)\rightarrow I_{2}(s)\]

Step 2: Balance elements other than H and O. The oxidation reaction is already balanced in terms of species other than H and O, but the I in the reduction reaction needs to be balanced. Since there is 1 I on the left side and 2 I on the right side, a coefficient 2 should be added to the I on the left.

Oxidation Reaction: \[HNO_{2}(aq)\rightarrow NO(g)\]

Reduction Reaction: \[2I^{-}(aq)\rightarrow I_{2}(s)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 1 water molecule to the right while the reduction reaction does not contain any oxygens, thus does not need water molecules added.

Oxidation Reaction: \[HNO_{2}(aq)\rightarrow NO(g)+H_{2}O(l)\]

Reduction Reaction: \[2I^{-}(aq)\rightarrow I_{2}(s)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 1 proton to the left while the reduction reaction does not contain any hydrogens, thus does not need protons added.

Oxidation Reaction: \[H^{+}(aq)+HNO_{2}(aq)\rightarrow NO(g)+H_{2}O(l)\]

Reduction Reaction: \[2I^{-}(aq)\rightarrow I_{2}(s)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 1+ charge on the left side and 0 charge on the right side. To balance the reaction we need to add 1 electrons to the left side.

Oxidation Reaction: \[e^{-}+H^{+}(aq)+HNO_{2}(aq)\rightarrow NO(g)+H_{2}O(l)\]

The reduction reaction has a (1-) + (1-) = 2- charge on the left side and 0 charge on the right side. To balance the equation we need to add 2 electrons to the right side.

Reduction Reaction: \[2I^{-}(aq)\rightarrow I_{2}(s)+2e^{-}\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 1 electon and the reduction reaction has 2 electrons. Thus the oxidation reaction must be multiplied by 2 and the reduction reaction by 1.

Oxidation Reaction: \[2[e^{-}+H^{+}(aq)+HNO_{2}(aq)\rightarrow NO(g)+H_{2}O(l)]\Rightarrow 2e^{-}+2H^{+}(aq)+2HNO_{2}(aq)\rightarrow 2NO(g)+2H_{2}O(l)\]

Reduction Reaction: \[1[2I^{-}(aq)\rightarrow I_{2}(s)+2e^{-}]\Rightarrow 2I^{-}(aq)\rightarrow I_{2}(s)+2e^{-}\]

Step 7: Add the reactions and cancel out common terms.

\[2e^{-}+2H^{+}(aq)+2HNO_{2}(aq)+2I^{-}(aq)\rightarrow 2NO(g)+2H_{2}O(l)+I_{2}(s)+2e^{-}\]

The electrons cancel out. Since this reaction takes place in a acidic solution, this is the complete balanced chemical equation of the original redox reaction.

\(2H^{+}(aq)+2HNO_{2}(aq)+2I^{-}(aq)\rightarrow 2NO(g)+2H_{2}O(l)+I_{2}(s)\)

C) \(N_{2}O(g) + ClO^{−}(aq) \rightarrow Cl^{−}(aq) + NO^{2−}(aq)\) in basic solution (Fixed MathJax)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing nitrogen because N goes from a 1+ charge to a 3+ charge. The reduction reaction is the one containing chlorine because Cl goes from a 1+ charge to a 1- charge.

Oxidation Reaction: \[N_{2}O(g)\rightarrow NO_{2}^{-}(aq)\]

Reduction Reaction: \[ClO^{-}(aq)\rightarrow Cl^{-}(aq)\]

Step 2: Balance elements other than H and O. The reduction reaction is already balanced in terms of species other than H and O, but the N in the reduction reaction needs to be balanced. Since there is 2 N on the left side and 1 N on the right side, a coefficient 2 should be added to the nitrogen species on the right.

Oxidation Reaction: \[N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)\]

Reduction Reaction: \[ClO^{-}(aq)\rightarrow Cl^{-}(aq)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 3 water molecules to the left while the reduction reaction needs to be balanced by adding 1 water molecule to the right.

Oxidation Reaction: \[3H_{2}O(l)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)\]

Reduction Reaction: \[ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 6 protons to the right while the reduction reaction needs to be balanced by adding 2 protons to the left.

Oxidation Reaction: \[3H_{2}O(l)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)+6H^{+}(aq)\]

Reduction Reaction: \[2H^{+}(aq)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 0 charge on the left side and (2-) + (6+)= 4+ charge on the right side. To balance the reaction we need to add 4 electrons to the right side.

Oxidation Reaction: \[3H_{2}O(l)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)+6H^{+}(aq)+4e^{-}\]

The reduction reaction has a (2+) + (1-) = 1+ charge on the left side and -1 charge on the right side. To balance the equation we need to add 2 electron to the left side.

Reduction Reaction: \[2e^{-}+2H^{+}(aq)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 4 electrons and the reduction reaction has 2 electrons. Thus the oxidation reaction must be multiplied by 1 and the reduction reaction by 2.

Oxidation Reaction: \[1[3H_{2}O(l)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)+6H^{+}(aq)+4e^{-}]\Rightarrow 3H_{2}O(l)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)+6H^{+}(aq)+4e^{-}\]

Reduction Reaction: \[2[2e^{-}+2H^{+}(aq)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)]\Rightarrow 4e^{-}+4H^{+}(aq)+2ClO^{-}(aq)\rightarrow 2Cl^{-}(aq)+2H_{2}O(l)\]

Step 7: Add the reactions and cancel out common terms.

\[4e^{-}+4H^{+}(aq)+2ClO^{-}(aq)+3H_{2}O(l)+N_{2}O(g)\rightarrow 2Cl^{-}(aq)+2H_{2}O(l)+2NO_{2}^{-}(aq)+6H^{+}(aq)+4e^{-}\]

The electrons cancel out as well as 4 protons and 2 water molecules

\[2ClO^{-}(aq)+H_{2}O(l)+N_{2}O(g)\rightarrow 2Cl^{-}(aq)+2H_{2}O(l)+2NO_{2}^{-}(aq)+2H^{+}(aq)\]

Step 8: Since this reaction takes place in basic solutions, the final answer must contain hydroxide. For this step we can add on hydroxide to both sides of the equation to balance the protons. There is 2 net proton in this equation so we add 2 hydroxide to both sides.

\[2OH^{-}(aq)+2ClO^{-}(aq)+H_{2}O(l)+N_{2}O(g)\rightarrow 2Cl^{-}(aq)+2NO_{2}^{-}(aq)+2H^{+}(aq)+2OH^{-}(aq)\]

Step 9: Combine the hydroxide and proton ions present on the same side to form water.

\[2OH^{-}(aq)+2ClO^{-}(aq)+H_{2}O(l)+N_{2}O(g)\rightarrow 2Cl^{-}(aq)+2NO_{2}^{-}(aq)+2H_{2}O(l)\]

Step 10: Cancel common terms to finalize the complete balanced chemical equation of the original redox reaction

\(2OH^{-}(aq)+2ClO^{-}(aq)+N_{2}O(g)\rightarrow 2Cl^{-}(aq)+H_{2}O(l)+2NO_{2}^{-}(aq)\)

D) \(Br_{2}(l) \rightarrow Br^{−}(aq) + BrO_{3}^{−}(aq)\) in basic solution (Fixed MathJax)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing oxygen because Br goes from 0 charge to a 5+ charge. The reduction reaction is the other one because Br goes from a 0 charge to a 1- charge.

Oxidation Reaction: \[Br_{2}(l)\rightarrow BrO^{-}_{3}(aq)\]

Reduction Reaction: \[Br_{2}(l)\rightarrow Br^{-}(aq)\]

Step 2: Balance elements other than H and O. Br in both reactions needs to be balanced. Since in the reduction reaction there are 2 Br on the left side and 1 Br on the right side, a coefficient 2 should be added to the bromine species on the right. The same balancing needs to happen in the oxidation reaction. Since there are 2 Br on the left side and 1 Br on the right side, a coefficient 2 should be added to the bromine species on the right.

Oxidation Reaction: \[Br_{2}(l)\rightarrow 2BrO^{-}_{3}(aq)\]

Reduction Reaction: \[Br_{2}(l)\rightarrow 2Br^{-}(aq)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 6 water molecules to the left while the reduction reaction does not have oxygen, thus it does not need any water added to it.

Oxidation Reaction: \[6H_{2}O(l)+Br_{2}(l)\rightarrow 2BrO^{-}_{3}(aq)\]

Reduction Reaction: \[Br_{2}(l)\rightarrow 2Br^{-}(aq)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 12 protons to the right while the reduction reaction does not have hydrogen, thus does not need any protons added to it.

Oxidation Reaction: \[6H_{2}O(l)+Br_{2}(l)\rightarrow 2BrO^{-}_{3}(aq)+12H^{+}(aq)\]

Reduction Reaction: \[Br_{2}(l)\rightarrow 2Br^{-}(aq)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has 0 charge on the left side and (2-) + (12+)= 10+ charge on the right side. To balance the reaction we need to add 10 electrons to the right side.

Oxidation Reaction: \[6H_{2}O(l)+Br_{2}(l)\rightarrow 2BrO^{-}_{3}(aq)+12H^{+}(aq)+10e^{-}\]

The reduction reaction has 0 charge on the left side and a 2- charge on the right side. To balance the equation we need to add 2 electron to the left side.

Reduction Reaction: \[2e^{-}+Br_{2}(l)\rightarrow 2Br^{-}(aq)\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 10 electrons and the reduction reaction has 2 electrons. Thus the oxidation reaction must be multiplied by 1 and the reduction reaction by 5.

Oxidation Reaction: \[1[6H_{2}O(l)+Br_{2}(l)\rightarrow 2BrO^{-}_{3}(aq)+12H^{+}(aq)+10e^{-}]\Rightarrow 6H_{2}O(l)+Br_{2}(l)\rightarrow 2BrO^{-}_{3}(aq)+12H^{+}(aq)+10e^{-}\]

Reduction Reaction: \[5[2e^{-}+Br_{2}(l)\rightarrow 2Br^{-}(aq)]\Rightarrow 10e^{-}+5Br_{2}(l)\rightarrow 10Br^{-}(aq)\]

Step 7: Add the reactions and cancel out common terms.

\[10e^{-}+5Br_{2}(l)+6H_{2}O(l)+Br_{2}(l)\rightarrow 10Br^{-}(aq)+2BrO^{-}_{3}(aq)+12H^{+}(aq)+10e^{-}\]

The electrons cancel out and the Br2 molecules combine

\[6Br_{2}(l)+6H_{2}O(l)\rightarrow 10Br^{-}(aq)+2BrO^{-}_{3}(aq)+12H^{+}(aq)\]

Step 8: Since this reaction takes place in basic solutions, the final answer must contain hydroxide. For this step we can add on hydroxide to both sides of the equation to balance the protons. There is 12 net proton in this equation so we add 12 hydroxide to both sides.

\[12OH^{-}(aq)+6Br_{2}(l)+6H_{2}O(l)\rightarrow 10Br^{-}(aq)+2BrO^{-}_{3}(aq)+12H^{+}(aq)+12OH^{-}(aq)\]

Step 9: Combine the hydroxide and proton ions present on the same side to form water.

\[12OH^{-}(aq)+6Br_{2}(l)+6H_{2}O(l)\rightarrow 10Br^{-}(aq)+2BrO^{-}_{3}(aq)+12H_{2}O(l)\]

Step 10: Cancel common terms to finalize the complete balanced chemical equation of the original redox reaction

\[12OH^{-}(aq)+6Br_{2}(l)\rightarrow 10Br^{-}(aq)+2BrO^{-}_{3}(aq)+6H_{2}O(l)\]

All coefficients are divisible by 2 so divide the equation by 2 to get the most simplified form

\(6OH^{-}(aq)+3Br_{2}(l)\rightarrow 5Br^{-}(aq)+BrO^{-}_{3}(aq)+3H_{2}O(l)\)

E) \(Cl(CH_{2})_{2}OH(aq) + K_{2}Cr_{2}O_{7}(aq) \rightarrow ClCH_{2}CO_{2}H(aq) + Cr^{3+}(aq)\) in acidic solution (Fixed Mathjax)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing carbon because C goes from a -1 charge to a +1 charge. The reduction reaction is the one containing chromium because Cr goes from a 12+ charge to a 3+ charge.

Oxidation Reaction: \[Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)\]

Reduction Reaction: \[K_{2}Cr_{2}O_{7}(aq)\rightarrow Cr^{3+}(aq)\]

Step 2: Balance elements other than H and O. The oxidation reaction is already balanced in terms of species other than H and O, but the Cr in the reduction reaction needs to be balanced. Since there is 2Cr on the left side and 1 Cr on the right side, a coefficient 2 should be added to the Cr on the right. Also, since there are 2 potassiums on the right side of the reduction reaction and none on the left side, we can add 2 potassiums to the right side to balance it out.

Oxidation Reaction: \[Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)\]

Reduction Reaction: \[K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 1 water molecule on the left side while the reduction reaction needs to be balanced by adding 7 water molecules on the right side.

Oxidation Equation: \[H_{2}O(l)+Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)\]

Reduction Reaction: \[K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 4 protons to the right side while the reduction reaction needs to be balanced by adding 14 protons to the left side.

Oxidation Equation: \[H_{2}O(l)+Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)+4H^{+}(aq)\]

Reduction Reaction: \[14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 0 charge on the left side and a 4+ charge on the right side. To balance the reaction we need to add 4 electrons to the right side.

Oxidation Reaction: \[H_{2}O(l)+Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-}\]

The reduction reaction has a 14+ charge on the left side and (6+) + (2+)= 8+ charge on the right side. To balance the equation we need to add 6 electrons to the left side.

Reduction Reaction: \[6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 4 electrons and the reduction reaction has 6 electrons. Thus the oxidation reaction must be multiplied by 3 and the reduction reaction by 2.

Oxidation Reaction: \[3[H_{2}O(l)+Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-}]\Rightarrow 3H_{2}O(l)+3Cl(CH_{2})_{2}OH(aq)\rightarrow 3ClCH_{2}CO_{2}H(aq)+12H^{+}(aq)+12e^{-}\]

Reduction Reaction: \[2[6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)]\Rightarrow 12e^{-}+28H^{+}(aq)+2K_{2}Cr_{2}O_{7}(aq)\rightarrow 4Cr^{3+}(aq)+4K^{+}(aq)+14H_{2}O(l)\]

Step 7: Add the reactions and cancel out common terms.

\[12e^{-}+28H^{+}(aq)+2K_{2}Cr_{2}O_{7}(aq)+3H_{2}O(l)+3Cl(CH_{2})_{2}OH(aq)\rightarrow 4Cr^{3+}(aq)+4K^{+}(aq)+14H_{2}O(l)+3ClCH_{2}CO_{2}H(aq)+12H^{+}(aq)+12e^{-}\]

The electrons cancel out as well as 12 protons and 3 water molecules. Since this reaction takes place in a acidic solution, this is the complete balanced chemical equation of the original redox reaction.

\(16H^{+}(aq)+2K_{2}Cr_{2}O_{7}(aq)+3Cl(CH_{2})_{2}OH(aq)\rightarrow 4Cr^{3+}(aq)+4K^{+}(aq)+11H_{2}O(l)+3ClCH_{2}CO_{2}H(aq)\)

Q20.4.16

Write a balanced chemical equation for each redox reaction:

a) In an acidic solution: \(I^{-}(aq)+HClO^{2}(aq)\rightarrow IO_{3}^{-}(aq)+Cl_{2}(g)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing chlorine because Cl goes from a +3 charge to a 0 charge. The reduction reaction is the one containing iodine because I goes from a 1- charge to a 5+ charge.

Oxidation Reaction: \[HClO^{2}(aq)\rightarrow Cl_{2}(g)\]

Reduction Reaction: \[I^{-}(aq)\rightarrow IO_{3}^{-}(aq)\]

Step 2: Balance elements other than H and O. The reduction reaction is already balanced in terms of species other than H and O, but the Cl in the reduction reaction needs to be balanced. Since there are 2 Cl on the right side and 1 Cl on the left side, a coefficient 2 should be added to the Cl species on the left.

Oxidation Reaction: \[2HClO^{2}(aq)\rightarrow Cl_{2}(g)\]

Reduction Reaction: \[I^{-}(aq)\rightarrow IO_{3}^{-}(aq)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 4 water molecules on the right side while the reduction reaction needs to be balanced by adding 3 water molecules on the left side.

Oxidation Equation: \[2HClO^{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)\]

Reduction Reaction: \[3H_{2}O(l)+I^{-}(aq)\rightarrow IO_{3}^{-}(aq)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 6 protons to the left side while the reduction reaction needs to be balanced by adding 6 protons to the right side.

Oxidation Equation: \[6H^{+}(aq)+2HClO^{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)\]

Reduction Reaction: \[3H_{2}O(l)+I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 6+ charge on the left side and a 0 charge on the right side. To balance the reaction we need to add 6 electrons to the left side.

Oxidation Reaction: \[6e^{-}+6H^{+}(aq)+2HClO^{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)\]

The reduction reaction has a 1- charge on the left side and (1-) + (6+)= 5+ charge on the right side. To balance the equation we need to add 6 electrons to the right side.

Reduction Reaction: \[3H_{2}O(l)+I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\]

Step 6: Scale the reactions so that the electrons are equal. In this problem, the electrons are already equal, so the reactions do not need to be scaled.

Step 7: Add the reactions and cancel out common terms.

\[6e^{-}+6H^{+}(aq)+2HClO^{2}(aq)+3H_{2}O(l)+I^{-}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)+IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\]

The electrons cancel out as well as 6 protons and 3 water molecules. Since this reaction takes place in a acidic solution, this is the complete balanced chemical equation of the original redox reaction.

\(2HClO^{2}(aq)+I^{-}(aq)\rightarrow Cl_{2}(g)+H_{2}O(l)+IO_{3}^{-}(aq)\)

b) In an acidic solution: \(Cr^{2+}(aq)+O_{2}(g)\rightarrow Cr^{3+}(aq)+H_{2}O(l)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing oxygen because O goes from a 0 charge to a 2- charge. The reduction reaction is the one containing Cr because chromium goes from a 2+ charge to a 3+ charge.

Oxidation Reaction: \[O_{2}(g)\rightarrow H_{2}O(l)\]

Reduction Reaction: \[Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)\]

Step 2: Balance elements other than H and O. But in this case, the oxidation reaction is comprised of only O and H so we need to balance the amount of oxygen on both sides without adding water. Since there are 2 oxygen on the left side and one on the right, a coefficient 2 should be added to the water molecule on the right side. The reduction reaction is already balanced in terms of species other than H and O.

Oxidation Reaction: \[O_{2}(g)\rightarrow 2H_{2}O(l)\]

Reduction Reaction: \[Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)\]

Step 3: Add water to balance the oxygen in each reaction, but the oxidation reaction had its oxygen balanced in step 2 and the reduction reaction has no oxygen needed to balance with water molecules. Move on to the next step.

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 4 protons to the left side while the reduction reaction has no hydrogen needed to balance with protons.

Oxidation Equation: \[4H^{+}(aq)+O_{2}(g)\rightarrow 2H_{2}O(l)\]

Reduction Reaction: \[Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 4+ charge on the left side and a 0 charge on the right side. To balance the reaction we need to add 4 electrons to the left side.

Oxidation Reaction: \[4e^{-}+4H^{+}(aq)+O_{2}(g)\rightarrow 2H_{2}O(l)\]

The reduction reaction has a 2+ charge on the left side and a 3+ charge on the right side. To balance the equation we need to add 1 electron to the right side.

Reduction Reaction: \[Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)+e^{-}\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 4 electrons and the reduction reaction has 1 electron. Thus the oxidation reaction must be multiplied by 1 and the reduction reaction by 4.

Oxidation Reaction: \[1[4e^{-}+4H^{+}(aq)+O_{2}(g)\rightarrow 2H_{2}O(l)]\Rightarrow 4e^{-}+4H^{+}(aq)+O_{2}(g)\rightarrow 2H_{2}O(l)\]

Reduction Reaction: \[4[Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)+e^{-}]\Rightarrow 4Cr^{2+}(aq)\rightarrow 4Cr^{3+}(aq)+4e^{-}\]

Step 7: Add the reactions and cancel out common terms.

\[4e^{-}+4H^{+}(aq)+O_{2}(g)+4Cr^{2+}(aq)\rightarrow 2H_{2}O(l)+4Cr^{3+}(aq)+4e^{-}\]

The electrons cancel out. Since this reaction takes place in a acidic solution, this is the complete balanced chemical equation of the original redox reaction.

\(4H^{+}(aq)+O_{2}(g)+4Cr^{2+}(aq)\rightarrow 2H_{2}O(l)+4Cr^{3+}(aq)\)

c) In a basic solution: \(CrO^{-}_{2}(aq)+ClO^{-}(aq)\rightarrow CrO^{2-}_{4}(aq)+Cl^{-}(aq)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing chlorine because Cl goes from a 1+ charge to a 1- charge. The reduction reaction is the one containing Cr because chromium goes from a 3+ charge to a 4+ charge.

Oxidation Reaction: \[ClO^{-}(aq)\rightarrow Cl^{-}(aq)\]

Reduction Reaction: \[CrO^{-}_{2}(aq)\rightarrow CrO^{2-}_{4}(aq)\]

Step 2: Balance elements other than H and O. In this problem, everything but H and O is already balanced so we can move on to the next step.

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 1 water molecule on the right side while the reduction reaction needs to be balanced by adding 2 water molecules on the left side.

Oxidation Equation: \[ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

Reduction Reaction: \[2H_{2}O(l)+CrO^{-}_{2}(aq)\rightarrow CrO^{2-}_{4}(aq)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 2 protons to the left side while the reduction reaction needs to be balanced by adding 4 protons to the right side.

Oxidation Equation: \[2H^{+}(aq)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

Reduction Reaction: \[2H_{2}O(l)+CrO^{-}_{2}(aq)\rightarrow CrO^{2-}_{4}(aq)+4H^{+}(aq)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 1- charge on the left side and a 1+ charge on the right side. To balance the reaction we need to add 2 electrons to the left side.

Oxidation Reaction: \[2e^{-}+2H^{+}(aq)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

The reduction reaction has a 1- charge on the left side and (2-) + (4+) = 2+ charge on the right side. To balance the equation we need to add 3 electrons to the right side.

Reduction Reaction: \[2H_{2}O(l)+CrO^{-}_{2}(aq)\rightarrow CrO^{2-}_{4}(aq)+4H^{+}(aq)+3e^{-}\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 2 electrons and the reduction reaction has 3 electrons. Thus the oxidation reaction must be multiplied by 3 and the reduction reaction by 2.

Oxidation Reaction: \[3[2e^{-}+2H^{+}(aq)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+H_{2}O(l)]\Rightarrow 6e^{-}+6H^{+}(aq)+3ClO^{-}(aq)\rightarrow 3Cl^{-}(aq)+3H_{2}O(l)\]

Reduction Reaction: \[2[2H_{2}O(l)+CrO^{-}_{2}(aq)\rightarrow CrO^{2-}_{4}(aq)+4H^{+}(aq)+3e^{-}]\Rightarrow 4H_{2}O(l)+2CrO^{-}_{2}(aq)\rightarrow 2CrO^{2-}_{4}(aq)+8H^{+}(aq)+6e^{-}\]

Step 7: Add the reactions and cancel out common terms.

\[6e^{-}+6H^{+}(aq)+3ClO^{-}(aq)+4H_{2}O(l)+2CrO^{-}_{2}(aq)\rightarrow 3Cl^{-}(aq)+3H_{2}O(l)+2CrO^{2-}_{4}(aq)+8H^{+}(aq)+6e^{-}\]

The electrons cancel out as well as 6 protons and 3 water molecules.

\[3ClO^{-}(aq)+H_{2}O(l)+2CrO^{-}_{2}(aq)\rightarrow 3Cl^{-}(aq)+2CrO^{2-}_{4}(aq)+2H^{+}(aq)\]

Step 8: Since this reaction takes place in basic solutions, the final answer must contain hydroxide. For this step we can add on hydroxide to both sides of the equation to balance the protons. There is 2 net protons in this equation so we add 2 hydroxide to both sides.

\[2OH^{-}(aq)+3ClO^{-}(aq)+H_{2}O(l)+2CrO^{-}_{2}(aq)\rightarrow 3Cl^{-}(aq)+2CrO^{2-}_{4}(aq)+2H^{+}(aq)+2OH^{-}(aq)\]

Step 9: Combine the hydroxide and proton ions present on the same side to form water.

\[2OH^{-}(aq)+3ClO^{-}(aq)+H_{2}O(l)+2CrO^{-}_{2}(aq)\rightarrow 3Cl^{-}(aq)+2CrO^{2-}_{4}(aq)+2H_{2}O(l)\]

Step 10: Cancel common terms to finalize the complete balanced chemical equation of the original redox reaction

\(2OH^{-}(aq)+3ClO^{-}(aq)+2CrO^{-}_{2}(aq)\rightarrow 3Cl^{-}(aq)+2CrO^{2-}_{4}(aq)+H_{2}O(l)\)

d) In an acidic solution: \(S(s)+HNO_{2}(aq)\rightarrow H_{2}SO_{3}(aq)+N_{2}O(g)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing nitrogen because N goes from a 3+ charge to a 1+ charge. The reduction reaction is the one containing sulfur because S goes from a 0 charge to a 4+ charge.

Oxidation Reaction: \[HNO_{2}(aq)\rightarrow N_{2}O(g)\]

Reduction Reaction: \[S(s)\rightarrow H_{2}SO_{3}(aq)\]

Step 2: Balance elements other than H and O. Since there are 2 nitrogen on the right side and 1 on the left side, a coefficient 2 should be added to the nitrogen species on the left side. The reduction reaction is already balanced in terms of species other than H and O.

Oxidation Reaction: \[2HNO_{2}(aq)\rightarrow N_{2}O(g)\]

Reduction Reaction: \[S(s)\rightarrow H_{2}SO_{3}(aq)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 3 water molecules on the right side while the reduction reaction needs to be balanced by adding 3 water molecules on the left side.

Oxidation Reaction: \[2HNO_{2}(aq)\rightarrow N_{2}O(g)+3H_{2}O(l)\]

Reduction Reaction: \[3H_{2}O(l)+S(s)\rightarrow H_{2}SO_{3}(aq)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 4 protons to the left side while the reduction reaction needs to be balanced by adding 4 protons to the right side.

Oxidation Equation: \[4H^{+}(aq)+2HNO_{2}(aq)\rightarrow N_{2}O(g)+3H_{2}O(l)\]

Reduction Reaction: \[3H_{2}O(l)+S(s)\rightarrow H_{2}SO_{3}(aq)+4H^{+}(aq)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 4+ charge on the left side and a 0 charge on the right side. To balance the reaction we need to add 4 electrons to the left side.

Oxidation Reaction: \[4e^{-}+4H^{+}(aq)+2HNO_{2}(aq)\rightarrow N_{2}O(g)+3H_{2}O(l)\]

The reduction reaction has 0 charge on the left side and a 4+ charge on the right side. To balance the equation we need to add 4 electrons to the right side.

Reduction Reaction: \[3H_{2}O(l)+S(s)\rightarrow H_{2}SO_{3}(aq)+4H^{+}(aq)+4e^{-}\]

Step 6: Scale the reactions so that the electrons are equal. In this problem, the electrons are already equal, so the reactions do not need to be scaled.

Step 7: Add the reactions and cancel out common terms.

\[4e^{-}+4H^{+}(aq)+2HNO_{2}(aq)+3H_{2}O(l)+S(s)\rightarrow N_{2}O(g)+3H_{2}O(l)+H_{2}SO_{3}(aq)+4H^{+}(aq)+4e^{-}\]

The electrons cancel out as well as 4 protons and 3 water molecules. Since this reaction takes place in a acidic solution, this is the complete balanced chemical equation of the original redox reaction.

\(2HNO_{2}(aq)+S(s)\rightarrow N_{2}O(g)+H_{2}SO_{3}(aq)\)

e) In an acidic solution: \(F(CH_{2})_{2}OH(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow FCH_{2}CO_{2}H(aq)+Cr^{3+}(aq)\)

Step 1: Separate the half reactions into their corresponding oxidation reaction, or reaction that loses electrons, and reduction reaction, or reaction that gains electrons. The oxidation reaction is the one containing carbon because C goes from a -1 charge to a +1 charge. The reduction reaction is the one containing chromium because Cr goes from a 12+ charge to a 3+ charge.

Oxidation Reaction: \[F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)\]

Reduction Reaction: \[K_{2}Cr_{2}O_{7}(aq)\rightarrow Cr^{3+}(aq)\]

Step 2: Balance elements other than H and O. The oxidation reaction is already balanced in terms of species other than H and O, but the Cr in the reduction reaction needs to be balanced. Since there is 2Cr on the left side and 1 Cr on the right side, a coefficient 2 should be added to the Cr on the right. Also, since there are 2 potassium on the right side of the reduction reaction and none on the left side, we can add 2 potassiums to the right side to balance it out.

Oxidation Reaction: \[F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)\]

Reduction Reaction: \[K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)\]

Step 3: Add water to balance the oxygen in each reaction. The oxidation reaction needs to be balanced by adding 1 water molecule on the left side while the reduction reaction needs to be balanced by adding 7 water molecules on the right side.

Oxidation Equation: \[H_{2}O(l)+F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)\]

Reduction Reaction: \[K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)\]

Step 4: Add protons to balance the hydrogen in each reaction. The oxidation reaction needs to be balanced by adding 4 protons to the right side while the reduction reaction needs to be balanced by adding 14 protons to the left side.

Oxidation Equation: \[H_{2}O(l)+F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)+4H^{+}(aq)\]

Reduction Reaction: \[14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)\]

Step 5: Balance the charge of the equations with electrons. The oxidation reaction has a 0 charge on the left side and a 4+ charge on the right side. To balance the reaction we need to add 4 electrons to the right side.

Oxidation Reaction: \[H_{2}O(l)+F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-}\]

The reduction reaction has a 14+ charge on the left side and (6+) + (2+)= 8+ charge on the right side. To balance the equation we need to add 6 electrons to the left side.

Reduction Reaction: \[6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)\]

Step 6: Scale the reactions so that the electrons are equal. The oxidation reaction has 4 electrons and the reduction reaction has 6 electrons. Thus the oxidation reaction must be multiplied by 3 and the reduction reaction by 2.

Oxidation Reaction: \[3[H_{2}O(l)+F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-}]\Rightarrow 3H_{2}O(l)+3F(CH_{2})_{2}OH(aq)\rightarrow 3FCH_{2}CO_{2}H(aq)+12H^{+}(aq)+12e^{-}\]

Reduction Reaction: \[2[6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{3+}(aq)+2K^{+}(aq)+7H_{2}O(l)]\Rightarrow 12e^{-}+28H^{+}(aq)+2K_{2}Cr_{2}O_{7}(aq)\rightarrow 4Cr^{3+}(aq)+4K^{+}(aq)+14H_{2}O(l)\]

Step 7: Add the reactions and cancel out common terms.

\[12e^{-}+28H^{+}(aq)+2K_{2}Cr_{2}O_{7}(aq)+3H_{2}O(l)+3F(CH_{2})_{2}OH(aq)\rightarrow 4Cr^{3+}(aq)+4K^{+}(aq)+14H_{2}O(l)+3FCH_{2}CO_{2}H(aq)+12H^{+}(aq)+12e^{-}\]

The electrons cancel out as well as 12 protons and 3 water molecules. Since this reaction takes place in a acidic solution, this is the complete balanced chemical equation of the original redox reaction.

\(16H^{+}(aq)+2K_{2}Cr_{2}O_{7}(aq)+3F(CH_{2})_{2}OH(aq)\rightarrow 4Cr^{3+}(aq)+4K^{+}(aq)+11H_{2}O(l)+3FCH_{2}CO_{2}H(aq)\)

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