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?

S17.3.5

In this problem, it is given that cadmium metal is oxidized to 1 M cadmium(II). Because only liquid solutions have molarity, we know that cadmium transforms from its metal state to its liquid state via an oxidation reaction. Secondly, the problem notes that an aluminum electrode is placed into a 1 M aluminum nitrate solution. Because galvanic cells have an oxidation half reaction and a reduction half reaction, we note that aluminum in its solid form as an electrode and in its liquid form compose the reduction half reaction. It's useful to write out the cell notation of the reaction to exactly see how elements transform, however it is not necessary for this problem. If we were to write it out, it would look like this:

\[Cd(s) | Cd^{2+}(1M) || Al^{3+}(1M) | Al(s)\]

Cadmium is converted to its liquid phase, and because the electrodes must be placed at the ends of the cell notation, we see that aluminum goes from its liquid state to solid state. A single line represents a phase change while a double line represents the separation of the anode (oxidation) and cathode (reduction) half-reactions.

A quicker way to approach this is to realize that cadmium metal has an oxidation state of 0 and aluminum in aluminum nitrate has an oxidation state of +3. Then, because this must be the reduction half reaction, we write that aluminum(III) is reduced to Aluminum(0). Because this question asks for the cell potential of the overall reaction at standard conditions, we must use a Standard Reduction cell Potential chart to locate the cell potentials of the half reactions. With all of this information found, written out it looks like:

\[Cd(s)\rightarrow Cd^{2+}(aq) + 2e^{-}; E°_{anode}=-0.403 Volts\]
\[Al^{3+}(aq)+ 3e^{-}\rightarrow Al(s); E°_{cathode}=-1.66 Volts\]

To determine the overall reaction, we must combine the oxidation half and reduction half equations through balancing the electrons with stoichiometry.

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

Combining them and cancelling out the electrons, we then get this as the overall reaction:

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

Next, to find the overall cell potential, an equation where the oxidation half reaction potential is subtracted from the reduction half-reaction potential is used. Plugging in the necessary variables, we find:

\[E°_{cell}=E°_{cathode}-E°_{anode}= -1.66 V -(-0.403 V)\]
\[E°_{cell}=-1.257 V\]

Note that the cell potentials of the half reactions are unaffected by a reversal of the reaction or by the stoichiometry that is needed to combine these two half-reactions. This is because Standard Reduction Potentials account for such variations.

Because cell potential and gibbs energy have an inverse relationship when it comes to charges. If the cell potential is positive for a reaction, then the gibbs energy associated with that reaction is negative, and vice verca. Because the cell potential is negative here, the gibbs energy for this reaction must be positive. Further, because a reaction is spontaneous for \(Delta G< 0\) and nonspontaneous for \(Delta G > 0\), this reaction must be nonspontaneous because it falls under the latter restriction.

Q19.1.24

Predict which will be more stable, [CrO₄]²⁻ or [WO₄]²⁻, and explain.

S19.1.24

Analyzing this problem, it's cleat that there are many similarities. Both complexes have four bonds to a central metal atom, and both complexes have a -2 charge. Because Oxygen has a -2 oxidation state and there are four oxygen atoms in each complex, and because the overall charge for both complexes is -2, we calculate that both chromium and tungsten have an oxidation state of +6. The equation used is: x + 4(-2) = -2, where x is the charge of both metal atoms. This is +6.

Next, we analyze the differences of the central metal atoms. The first complex has chromium, and the second has tungsten. Looking at a periodic table, chromium and tungsten are located in the 3d orbital and 5d orbital respectively. This information is the key to conceptually solving this problem. Elements in higher orbitals have a greater amount of energy. This larger energy causes bonds to be longer and therefore weaker, as bond length is inversely proportional to atomic attraction.

In conclusion, because tungsten is in a higher d orbital than chromium, it has more energy and thus forms weaker and more unstable bonds. This allows [CrO₄]²⁻ to be the more stable complex.

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?

S12.4.17

The problem tells us that this reaction is a first order process, which means we want to use the first order process integration equation to determine the age of the sample, \[[A]=[A]_{0}e^{-kt}\]

We know that 1/64 of the initial dose, [A]₀ remains, and t is our variable. However, in order to solve for t, we have to know what k is. Because this is a first order process, we know that \[t_{1/2}=\frac{ln2}{k}\]

Rearranging the equation to solve for k, we get k = ln2/ t_1/2 . Plugging in the half life, 42 days into this equation, we get k = ln2/42, where k = 0.0165.

Now that we only have one variable, t, to solve for in the integrated formula, we plug in all known information and solve for the age of the sample.

\[\frac{1}{64} [A]_{0}=[A]_{0}e^{-0.0165t}\]
Dividing by [A]₀ and raising both sides to the power of ln, we get \[ln\frac{1}{64}=-0.0165t\]

Then dividing by -0.0165, we find that t = 252

The age of the sample is 252 days.

Q21.3.1

Write a brief description or definition of each of the following:

1. nucleon
2. α particle
3. β particle
4. positron
5. γ ray
6. nuclide
7. mass number
8. atomic number

S21.3.1

1. A nucleon is any particle in a nucleus. Therefore a nucleon can be a proton or a neutron.

2. α is the Greek symbol that symbolizes an alpha particle when in a radioactive context. An alpha particle is a product of natural radioactivity and is a helium atom, with two protons, two neutrons, two electrons, a nuclear mass of 4, and an atomic number of 2. Alpha particles are a form of radiation that can not penetrate a few centimeters of air, a thin sheet of paper, human skin or anything with a greater thickness. It's formula is \(_{2}^{4}\textrm{He}\)

3. β is the Greek symbol beta, and is used to describe beta particles. Beta particles are also products of natural radioactivity and are high-speed electrons from an atom's nucleus.

If the beta particle involves the emission of a positron, it is noted by several forms:\(_{1}^{0}\beta^{+}; beta^{+}; _{1}^{0}e\)

If a beta particle involves the emission of an electron, it is noted as:\(_{-1}^{0}\beta^{-}; beta^{-}; _{-1}^{0}e\)

Beta particles cannot penetrate wood or plastic.

4. A positron is a positive beta particle, or an electron with a positive charge. It is denoted by \(_{1}^{0}\beta^{+}; beta^{+}; _{1}^{0}e\)

5. γ is the Greek symbol gamma, and describes electromagnetic radiation from the sun. Gamma rays are chargeless, massless, and cannot penetrate concrete or lead. It is denoted by \(_{0}^{0}\gamma\)

6. A nuclide is a single specific nucleus that consists of its number of protons and neutrons and its mass.

7. The mass number is the total mass of all protons and neutrons in a nucleus of an atom of a specific isotopic element.

8. The atomic number is the number that defines any element. It describes the number of protons in the nucleus of an atom, which is the single unchanging factor of an atom that determines its identity.

Q21.7.3

Given specimens uranium-232 (t1/2=68.9y) and uranium-233 (t1/2=159,200y) of equal mass, which one would have greater activity and why?

S21.7.3

Activity is the rate of decay of a radioactive sample of matter. The integrated activity equation is A = kN, where N is the number of atoms of a radioactive isotope left in a sample, and k is the decay rate constant. Because radioactivity is a first order process, we know that t_1/2 = ln2/k.

Looking at uranium-232, and the average half-life of 68.9, and solving for k, we find:

\[k = \frac{ln2}{68.9} = 0.0100\]

Looking at uranium-233, and the average half-life of 159,200, and solving for k, we find:

\[k = \frac{ln2}{159,200} = 4.35E-6\]

If we look back at the activity equation A = kN, and think about how each decay constant k would affect the activity, we see that the larger k would produce a larger activity. Therefore, because 0.0100 is the larger decay constant, uranium-232 has 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.

1. Cl₂(g) + H₂(g) → 2Cl⁻(aq) + 2H⁺(aq)
2. Br₂(aq) + Fe²⁺(aq) → 2Br⁻(aq) + Fe³⁺(aq)
3. Fe³⁺(aq) + Cd(s) → Fe²⁺(aq) + Cd²⁺(aq)

S20.4.13

1. For the reaction Cl₂(g) + H₂(g) → 2Cl⁻(aq) + 2H⁺(aq) to be balanced, there must be the same number of atoms of an element on either side of the reaction. We must look at each element individually.

We must split the reaction into half-reactions. We see that chlorine gas is converted to chlorine ions, and hydrogen gas is converted to hydrogen ions. Separating the reactions and balancing the charges with electrons, we find the half reactions. Then looking at a Standard Reduction Potential Table, we can find the cell potential for each half reaction. We must remember that SRPs are not affected by the reversal of a reaction or by stoichiometry.

\[Cl_{2}(g)+2e^{-}\rightarrow 2Cl^{-}(aq); E°_{reduction} = +1.36V\]

\[H_{2}(g)\rightarrow 2H^{+}(aq)+2e^{-}; E°_{oxidation} = +0.00V\]

To find the overall balanced equation, we must balance the electrons and combine the two half reactions. Noticing that both reactions already involve two electrons, we can directly combine the half reactions and cancel out the electrons to find the overall balanced equation:

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

To find the cell potential, we must remember that reduction occurs at the cathode and oxidation occurs at the anode. Further, we must use the equation specific to standard conditions to find the standard electrode potential of the cell.

\[E°_{cell}=E°_{cathode}-E°_{anode}=+1.36V - (+0.00V)\]

\[E°_{cell}=+1.36V\]

2. Following the same steps that were taken to solve the previous reaction, we must separate the half reactions and find the SRPs.

\[Br_{2}(aq)+2e^{-}\rightarrow 2Br^{-}(aq); E°_{reduction} = +1.0873V\]

\[Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+e^{-}; E°_{oxidation} = +0.771V\]

To find the overall balanced equation, we must balance the electrons and then combine the half reactions. We can balance the electrons by multiplying the oxidation half reaction by 2.

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

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

The balanced equation is:

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

To find the cell potential, use the cell potential equation for standard conditions.

\[E°_{cell}=E°_{cathode}-E°_{anode}=+1.0873V - (+0.7710V)\]

\[E°_{cell}=+0.3163V\]

3. Following the same steps that were taken to solve 1 and 2, we must separate the half reactions and find the SRPs.

\[Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq); E°_{reduction} = +0.771V\]

\[Cd(s)\rightarrow Cd^{2+}(aq)+2e^{-}; E°_{oxidation} = -0.403V\]

To find the overall balanced equation, we must balance the electrons and then combine the half reactions. We can balance the electrons by multiplying the reduction half reaction by 2.

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

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

The balanced equation is:

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

To find the cell potential, use the cell potential equation for standard conditions.

\[E°_{cell}=E°_{cathode}-E°_{anode}=+0.771V - (-0.403V\]

\[E°_{cell}=+1.174V\]

Q20.4.15

Write a balanced chemical equation for each redox reaction.

1. H₂PO₂⁻(aq) + SbO₂⁻(aq) → HPO₃²⁻(aq) + Sb(s) in basic solution
2. HNO₂(aq) + I⁻(aq) → NO(g) + I₂(s) in acidic solution
3. N₂O(g) + ClO⁻(aq) → Cl⁻(aq) + NO₂⁻(aq) in basic solution
4. Br₂(l) → Br⁻(aq) + BrO₃⁻(aq) in basic solution
5. Cl(CH₂)₂OH(aq) + K₂Cr₂O₇(aq) → ClCH₂CO₂H(aq) + Cr³⁺(aq) in acidic solution

S20.4.15

General Rules to Follow:

STEP 1: To balance a redox reaction, you first must separate the half reactions. Focus on one at a time

STEP 2: Balance all elements that are not hydrogen and oxygen

STEP 3: Balance oxygen with water

STEP 4: Balance hydrogen with H⁺

STEP 5: Balance the charges by adding electrons to the more positive side

STEP 6: If balancing in basic solution, then balance H⁺ with OH^- on both sides and combine to form water. Cancel water molecules accordingly

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms

1. H₂PO₂⁻(aq) + SbO₂⁻(aq) → HPO₃²⁻(aq) + Sb(s) in basic solution

STEP 1: Separate the half reactions. Focus on one at a time

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

STEP 2: Balance all elements that are not hydrogen and oxygen. Notice that phosphorous is balanced

STEP 3: Balance oxygen with water

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

STEP 4: Balance hydrogen with H⁺

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

STEP 5: Balance the charges by adding electrons to the more positive side

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

STEP 6: If balancing in basic solution, then balance H⁺ with OH^- on both sides and combine to form water. Cancel water molecules accordingly

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

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

Balance the other half reaction.

\[SbO_{2}^{-}(aq)\rightarrow Sb(s)\]

STEP 2: Balance all elements that are not hydrogen and oxygen. Notice that Antimony is balanced

STEP 3: Balance oxygen with water

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

STEP 4: Balance hydrogen with H⁺

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

STEP 5: Balance the charges by adding electrons to the more positive side

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

STEP 6: If balancing in basic solution, then balance H⁺ with OH^- on both sides and combine to form water. Cancel water molecules accordingly

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

\[3e^{-}+2H_{2}O(l)+SbO_{2}^{-}(aq)\rightarrow Sb(s)+4OH^{-}(aq)\]

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms

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

\[2(3e^{-}+2H_{2}O(l)+SbO_{2}^{-}(aq)\rightarrow Sb(s)+4OH^{-}(aq))\]

The overall reaction is

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

2. HNO₂(aq) + I⁻(aq) → NO(g) + I₂(s) in acidic solution

Following STEPS 1-5, the half reactions are

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

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

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

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

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

The overall reaction is

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

3. N₂O(g) + ClO⁻(aq) → Cl⁻(aq) + NO₂⁻(aq) in basic solution

Following STEPS 1-6, the half reactions are

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

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

Following STEP 7:

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

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

The overall reaction is

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

4. Br₂(l) → Br⁻(aq) + BrO₃⁻(aq) in basic solution

Following STEPS 1-6, the half reactions are

\[12OH^{-}(aq)+Br_{2}(l)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2}O(l)+11e^{-}\]

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

Following STEP 7:

\[2(12OH^{-}(aq)+Br_{2}(l)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2}O(l)+11e^{-})\]

\[11(Br_{2}(l)+2e^{-}\rightarrow 2Br^{-}(aq))\]

The overall reaction is

\[24OH^{-}(aq)+13Br_{2}(l)\rightarrow 4BrO_{3}^{-}(aq)+22Br^{-}(aq)+12H_{2}O(l)\]

5. Cl(CH₂)₂OH(aq) + K₂Cr₂O₇(aq) → ClCH₂CO₂H(aq) + Cr³⁺(aq) in acidic solution

Following STEPS 1-5, the half reactions are

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

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

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

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

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

The overall reaction is

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

Q20.4.16

Write a balanced chemical equation for each redox reaction.

1. I⁻(aq) + HClO₂(aq) → IO₃⁻(aq) + Cl₂(g) in acidic solution
2. Cr²⁺(aq) + O₂(g) → Cr³⁺(aq) + H₂O(l) in acidic solution
3. CrO₂⁻(aq) + ClO⁻(aq) → CrO₄²⁻(aq) + Cl⁻(aq) in basic solution
4. S(s) + HNO₂(aq) → H₂SO₃(aq) + N₂O(g) in acidic solution
5. F(CH₂)₂OH(aq) + K₂Cr₂O₇(aq) → FCH₂CO₂H(aq) + Cr³⁺(aq) in acidic solution

S20.4.16

General Rules to Follow:

STEP 1: To balance a redox reaction, you first must separate the half reactions. Focus on one at a time

STEP 2: Balance all elements that are not hydrogen and oxygen

STEP 3: Balance oxygen with water

STEP 4: Balance hydrogen with H⁺

STEP 5: Balance the charges by adding electrons to the more positive side

STEP 6: If balancing in basic solution, then balance H⁺ with OH^- on both sides and combine to form water. Cancel water molecules accordingly

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms

1. I⁻(aq) + HClO₂(aq) → IO₃⁻(aq) + Cl₂(g) in acidic solution

STEP 1: Separate the half reactions. Focus on one at a time

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

STEP 2: Balance all elements that are not hydrogen and oxygen. Note that iodine is balanced.

STEP 3: Balance oxygen with water

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

STEP 4: Balance hydrogen with H⁺
\[3H_{2}O(l)+ I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)\]

STEP 5: Balance the charges by adding electrons to the more positive side
\[3H_{2}O(l)+ I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\]

STEP 6: Skip. We are balancing in acidic solution

Balance the other half reaction.

\[HClO_{2}(aq)\rightarrow Cl_{2}(g)\]

STEP 2: Balance all elements that are not hydrogen and oxygen

\[2HClO_{2}(aq)\rightarrow Cl_{2}(g)\]

STEP 3: Balance oxygen with water

\[2HClO_{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)\]

STEP 4: Balance hydrogen with H⁺

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

STEP 5: Balance the charges by adding electrons to the more positive side

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

STEP 6: Skip. We are balancing in acidic solution

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms. Notice that the electrons are already equal.

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

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

The overall reaction is

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

2. Cr²⁺(aq) + O₂(g) → Cr³⁺(aq) + H₂O(l) in acidic solution

Following STEPS 1-5, the half reactions are

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

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

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

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

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

The overall reaction is:

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

3. CrO₂⁻(aq) + ClO⁻(aq) → CrO₄²⁻(aq) + Cl⁻(aq) in basic solution

Following STEPS 1-6, the half reactions are

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

Notice that the electrons are already balanced. Following STEP 7, and combining the two half reactions, the overall reaction is

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

4. S(s) + HNO₂(aq) → H₂SO₃(aq) + N₂O(g) in acidic solution

Notice that sulfur is balanced, while there is one more nitrogen in the products. There are also one more hydrogen and two more oxygens in the products. If we take all of these into account, we'll notice that adding one more HNO₂ balances out these differences.

The overall reaction is

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

5. F(CH₂)₂OH(aq) + K₂Cr₂O₇(aq) → FCH₂CO₂H(aq) + Cr³⁺(aq) in acidic solution

Following STEPS 1-5, the half reactions are

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

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

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

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

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

The overall reaction is

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