Extra Credit 3

Problem 17.1.3

For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

Explanation:

Reduction is the gain of electrons, while oxidation is the loss of electrons. An easy way to remember is using the acronym OIL RIG: Oxidation Is Losing, Reduction Is Gaining. In the half-reaction, when it is oxidized, the electrons will be on the product side, and for reduction, the electrons will be on the reactant side since they are gaining electrons. Another way to tell is the oxidation state of the element. Oxidation state is the charge that the ion possesses. For an element that is in its elemental state, such as O₂, it possesses an oxidation state of 0, meanwhile if it was H₂O, the oxidation state would change to -2 since the oxygen is attached to the two hydrogen will oxidation states of +1. When the oxidation state decreases, that means a reduction has occurred, and when it increases, an oxidation has occurred.

1. \(\mathrm{Fe^{3+}}\: +\: \mathrm{3e^{-}}\rightarrow \mathrm{Fe}\)

Answer: \(\mathrm{Fe^{3+}}\) is gaining electrons, hence why the oxidation number for Fe is reduced from 3+ to 0. Therefore, the half-reaction is going through reduction.

2. \(\mathrm{Cr\rightarrow Cr^{3+}\: +\: 3e^{-}}\)

Answer: \(\mathrm{Cr}\) is losing electrons, hence why the oxidation number for Cr is increased from 0 to 3+. Therefore, the half-reaction is going through oxidation.

3. \(\mathrm{MnO_{4}\,^{2-}\rightarrow MnO_{4}\, ^{-}\: +\: e^{-}}\)

Answer: \(\mathrm{MnO_{4}\,^{2-}}\) is losing electrons, hence why the oxidation number for Mn increased from +6 to +7. Therefore, the half-reaction is going through oxidation.

4. \(\mathrm{Li^{+}}\: +\: \mathrm{e^{-}}\rightarrow \mathrm{Li}\)

Answer: \(\mathrm{Li^{+}}\) is gaining electrons, hence why the oxidation number for Li decreased from +1 to 0. Therefore, the half-reaction is going through reduction.

Problem 19.1.1

Write the electron configurations for each of the following elements:

1. Sc

2. Ti

3. Cr

4. Fe

5. Ru

Explanation:

Electron configurations are used to show the arrangement of electrons around the nucleus. They are usually four orbitals in which the electrons fill, which are the s, p, d, and f orbitals. Each orbital has a distinct amount of subshells that can hold electrons. The s orbital has one subshell, the p has three, the d has five, and the f has seven. Each subshell can hold two electrons with opposite spins. When writing the electron configuration, you must use the periodic table to see where the element is, and how many electrons in each orbital it has. Groups 1 and 2 are s orbitals, group 13-18 are p orbitals, group 3-12 are d orbitals, and the bottom elements are usually f orbitals. In order to find the electron configuration, find where the element is on the periodic table, and in order to simplify your answer, you can use put the noble gas in brackets to signify the electron configuration of the noble gas, and continue on from there. Count how many electrons are in the s and d orbitals, and write it as a subscript.

Example:

Zn has the electron configuration of \(\mathrm{[Ar]4s^{2}3d^{10}}\)

1. Answer: \(\mathrm{[Ar]4s^{2}3d^{1}}\)
2. Answer: \(\mathrm{[Ar]4s^{2}3d^{2}}\)
3. Answer: \(\mathrm{[Ar]4s^{1}3d^{5}}\) The s-orbital has only one electron because the configuration is unstable, therefore to make it stable we must take an electron from the s-orbital to the d-orbital.
4. Answer: \(\mathrm{[Ar]4s^{2}3d^{6}}\)
5. Answer: \(\mathrm{[Kr]5s^{1}4d^{7}}\) The s-orbital has only one electron because the d⁶ configuration is unstable, therefore to make it stable we must take an electron from the s-orbital to the d-orbital.

Problem 19.2.3

Give the coordination number for each metal ion in the following compounds:

Recall what a coordination number is. Coordination number is the number of bonds the complex metal has in the entire complex. Coordination number is not the same as the number of ligands bonded to the metal because some ligands are bidentate. Bidentate means that the ligand donates two pairs of electrons to the central atom, but this does not mean that the coordination number is more for that ligand. Two examples of bidentate ligands are the oxalate ion, shortened to (ox) when being written, and ethylenediamine, shortened to (en) when being written.

1. [Co(CO₃)₃]³⁻ (note that CO₃²⁻ is bidentate in this complex)

Answer:Coordination number is 6 because CO₃²⁻ is a bidentate in this complex.

2. [Cu(NH₃)₄]²⁺

Answer:Coordination number is 4.

3.[Co(NH₃)₄Br₂]₂(SO₄)₃

Answer:Coordination number is 6.

4. [Pt(NH₃)₄][PtCl₄]

Answer:Coordination number is 4.

5. [Cr(en)₃](NO₃)₃

Answer:Coordination number is 6.

6. [Pd(NH₃)₂Br₂] (square planar)

Answer:Coordination number is 4.

7. K₃[Cu(Cl)₅]

Answer:Coordination number is 6.

8. [Zn(NH₃)₂Cl₂]

Answer:Coordination number is 4.

Problem 12.3.15

Nitrogen(II) oxide reacts with chlorine according to the equation:

\(2NO(g)\: +\: Cl_{2}(g)\rightarrow 2NOCl(g)\)

The following initial rates of reaction have been observed for certain reactant concentrations:

Question: What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl₂? What is the rate constant? What are the orders with respect to each reactant?

Answer:

Explanation:

In order to solve this, you must understand the relationship between the rate, the rate constant, and the concentrations. This relationship is described in the rate law equation.

\[rate = k[A]^{n}[B]^{m}\]

[A] and [B] are the concetrations that would be used in the equation, and in this case it would be NO and Cl₂.

When setting up the equation, you use the rate law ratio from the experiment, and set it equal to the ratio of the respective concentrations. Be careful not to put the exponents on the rate ratio, but rather on the concentration ratio!

Another reminder is to remember that the rate and concentrations are given in mol/L, which is the same as saying molarity, M.

Step 1: Set up rate law equation

\[rate = k[NO]^{n}[Cl_{2}]^{m}\]

Step 2: Find the order of each (m and n) reactant by using experimental values.

Finding n for \([NO]^{n}\)

Trials	Concentration of NO	Concentration of	Rate
Trial #1	0.5M	0.5M	1.14
Trial #2	1.00M	0.5M	4.56

\[(\frac{0.5}{1.00})^{n}(\frac{0.5}{0.5})^{m}=(\frac{1.14}{4.56})\]

\[(0.5)^{n}(1)^{m}=(0.25)\]

* \((1)^{m}\) is always going to equal to 1 no matter what the "m" value is.

Ex: \((1)^{10}=(1)^{20}=(1)^{15}=1\)

So we can just focus on tying to find "n." We get \((0.5)^{n}=(0.25)\)

Now we ask ourselves, what value "n" will give us (0.25)?

We can use basic algebra to solve for "n."

To do this, use the log formula, having log_0.5(0.25)

We get n=2

Finding m for \([Cl_{2}]^{m}\)

Trials	Concentration of NO	Concentration of	Rate
Trial #2	1.00M	0.5M	4.56
Trial #3	1.00M	1.00M	9.12

\[(\frac{1.00}{1.00})^{n}(\frac{0.5}{1.00})^{m}=(\frac{4.56}{9.12})\]

* Remember \((1)^{n}\) is always going to equal 1 despite the "n" value.

\[(\frac{0.5}{1.00})^{m}=(\frac{4.56}{9.12})\]

\((0.5)^{m}=(0.5)\)

Now we ask ourselves, what value "m" will give us (0.50)?

m=1

Step 3: Now that we have found the reaction order for each reactant, we can plug the "m" and "n" values into our rate law equation.

\[rate = k[NO]^{n}[Cl_{2}]^{m}\]

\[rate = k[NO]^{2}[Cl_{2}]^{1}\]

Step 4: Use any experimental trials to find the rate constant, k.

\[rate = k[NO]^{2}[Cl_{2}]^{1}\]

Trial	Concentration of NO	Concentration of	Rate
Trial #1	0.5M	0.5M	1.14

For this stage, pick one of the experiment and use that experiment's rate and concentrations.

\[1.14\frac{M}{h}=k(0.5M)^{2}(0.5M)^{1}\]

\[1.14\frac{M}{h}=k(0.125M^{3})\]

\[1.14\frac{M}{h}(\frac{1}{0.125M^{3}})=k\]

*Notice the units!!

\(k=9.12\frac{1}{hM^{2}}\) or \(k=9.12h^{-1}M^{-1}\)

Problem 12.6.7

Write the rate equation for each of the following elementary reactions:

Recall that elementary reactions are only one step and have a reaction order of one for each reactant. The rate equation usually involves the rate constant, k, and the reactants. Do not place the coefficients as exponents since the exponents are determined by experiment and not the equation itself!

1.\(O_{3}\rightarrow O_{2}\: +\: O\)

Answer: \(rate=k[O_{3}]\)

2. \(O_{3}\: +\: Cl\rightarrow O_{2}\: +\: ClO\)

Answer: \(rate=k[O_{3}][Cl]\)

3. \(ClO\: +\: O\rightarrow Cl\: +\: O_{2}\)

Answer: \(rate=k[ClO][O]\)

4. \(O_{3}\: +\: NO\rightarrow NO_{2}\: +\: O_{2}\)

Answer: \(rate=k[O_{3}][NO]\)

5. \(NO_{2}\: +\: O\rightarrow NO\: +\: O_{2}\)

Answer: \(rate=k[NO_{2}][O]\)

20.3.7

Copper(II) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous \(CuSO_{4}\) solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of \(ZnCl_{2}\)?

Write half-reactions to show the chemical changes that are occurring.

Answer:

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

This is an oxidation reaction because Zn(s) is losing two electrons and its oxidation number increases by 2.

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

This is a reduction reaction because \(Cu^{2+}(aq)\) is gaining two electrons hence why its oxidation number decreases by 2.

Now we go back to the question, what is happening?

Answer:

After writing the half reactions, we can tell that the two reactions are coupled and this type of reaction is called a redox reaction. A redox reaction is when a reduction reaction and oxidation reaction are occurring at the same time and the electrons lost from one reaction is used to reduce another reaction.

The copper(II) sulfate solution is bright blue in the water because it is hydrated and hexaaquacopper(II) ion is present. When a zinc metal is placed in the beaker, the zinc strip begins to erode because it is being oxidized (losing electrons). The electrons will reduce copper(II) sulfate and essentially there will be less copper ions in the solution so the solution's blue color will fade in color.

What will happen if a piece of copper metal is placed in a colorless aqueous solution of \(\mathbf{ZnCl_{2}}\)?

Answer:

There would be no change in the solution beaker because Cu(s) cannot be oxidized by \(Zn^{2+}\) according to the standard reduction potential table. This is because Cu is above Zn in the reduction potential table, meaning that Cu is a better oxidizing agent, and is able to reduce better than Zn, thus Zn cannot replace Cu.

20.5.18

In acidic solution, permanganate (MnO₄⁻) oxidizes Cl⁻ to chlorine gas, and MnO₄⁻ is reduced to Mn²⁺(aq).

A. Write the balanced chemical equation for this reaction.

Answer:

Write out your half reactions and balance

Reduction: \(MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)\)

Oxidation: \(Cl^{-}(aq)\rightarrow Cl_{2}(g)\)

i. For the reduction reaction:

Step 1: Balance out the oxygen by adding water to the opposite side of the reaction.

\[MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)\: +\: 4H_{2}O(l)\]

Step 2: Now the oxygen has been balanced on both sides of the equation. Balance the hydrogen by adding to the opposite side.

\[8H^{+}(aq)\: +\: MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)\: +\: 4H_{2}O(l)\]

Step 3: Now that we have finished balancing, we must check the charges on both sides of the equation. Both sides must be equal; if they aren't equal, we must balance it by adding .

\[8H^{+}(aq)\: +\: MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)\: +\: 4H_{2}O(l)\]

Left side: (+8) + (-1) = (+7)

Right side: (+2) + (0) = (+2)

Step 4: Try to get both sides to equal to (+2). In order to do that, we must add 5 electrons to the left hand side.

\[5e^{-}\: +\: 8H^{+}(aq)\: +\: MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)\: +\: 4H_{2}O(l)\]

*Make sure both sides are balanced in charge and in elements.

ii. For the oxidation reaction:

Step 1: Balance the equation.

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

Step 2: Ensure both sides of the equation have the same charge.

Left hand side: (2)(-1) = (-2)

Right hand side: (2)(0) = (0)

In order to balance both sides, we must add two electrons to the right hand side.

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

*Make sure both sides are balanced in charge and in elements.

Combine both half reactions and write the overall reaction equation.

Multiply each half reaction by a number that would allow the electrons on both sides to cancel out when combining the half reactions.

\[(5e^{-}\: +\: 8H^{+}(aq)\: +\: MnO_{4}^{-}(aq)\rightarrow Mn^{2+}(aq)\: +\: 4H_{2}O(l))\times 2\]

\[(2Cl^{-}(aq)\rightarrow Cl_{2}(g)\: +\: 2e^{-})\times 5\]

\[=10e^{-}\: +\: 16H^{+}(aq)\: +\: 2MnO_{4}^{-}(aq)\rightarrow 2Mn^{2+}(aq)\: +\: 8H_{2}O(l)\]

\[=10Cl^{-}(aq)\rightarrow 5Cl_{2}(g)\: +\: 10e^{-}\]

Overall Balanced Equation:

\[=10Cl^{-}(aq)\: +\: 16H^{+}(aq)\: +\: 2MnO_{4}^{-}(aq)\rightarrow 2Mn^{2+}(aq)\: +\: 8H_{2}O(l)\: +\: 5Cl_{2}(g)\]

B.Determine E°_cell

Answer:

Recall: \[E^{0}_{cell}= E^{0}_{cathode}-E^{0}_{anode}\]

Ask yourself: what type of reaction occurs in the cathode and the anode?

Mnemonic to help you memorize: RED CAT , AN OX

REDuction CAThode

ANode OXidation

From the question, it is given that (MnO₄⁻) is reduced to (Mn²⁺) so this reaction will happen at the cathode.

(Cl⁻) is oxidized to Cl(g) so this reaction will happen at the anode.

To get the ,V values we must refer back to our standard reduction potential table.

Reduction Half Reactions	,V
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)	+1.51
Cl2(g) + 2e- → 2Cl-(aq)	+1.358

With the given information:

\[E^{0}_{cell}= (+1.51)-(+1.358)\]

\[E^{0}_{cell}= 0.152\: V\]

C. Calculate the equilibrium constant.

Answer:

Recall the relationship between the equilibrium constant, K and the cell potential.

\[E^{0}_{cell}= \frac{RT}{nF}lnK\]

Step 1: List out each variable and write down what you know:

\[E^{0}_{cell}= 0.152\: V\]

R=8.314 J/(mol • K)

T = 298K

n= number of transferred electrons which is 10 (refer to when we combined half reactions)

F= Faraday's Constant = 96,486 J/(mol • V)

K= ?

Step 2: Plug in your known values

\[0.152V=\frac{(8.314Jmol^{-1}K^{-1})(298K)}{(10mol)(96486Jmol^{-1}V^{-1})}lnK\]

Step 3: Solve for your unknown and ensure that units are cancelling out and accounted for. The reason for this is because logs, including ln, and exponents are unitless.

\[K=5.10\times 10^{25}\]

Alternative Method:

An alternative equation to use so solve for K is:

\[E^{0}_{cell}=\frac{0.025693V}{n}lnK\]

at 298 K

=\(0.152=\frac{0.025693V}{10mol}lnK\)

=\(59.16=lnK\)

=\(e^{59.16}=K\)

\(\mathbf{K=4.93\times 10^{25}}\)

As you can see, the answers are almost the same.

21.4.19

Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of \(_{43}^{99}\textrm{Tc}\).

Answer:

Recall that radioactive decay is a first order. A characteristic trait of first order reactions is that the half-life does not depend on concentration.

Formula used: \(\lambda = \frac{0.693}{t_{1/2}}\)

Given: 6.0 h

Plug given(s) into the equation and solve for .

\(\lambda = \frac{0.693}{6.0h}=0.12h^{-1}\)