Chapter 17, Problem 74QAP

Consider the reaction low at 25°C:
$3{\text{SO}}_4{}^{2-}\left(aq\right)+12{\text{H}}^{+}\left(aq\right)+2\text{Cr}\left(s\right)\to 3{\text{SO}}_2\left(g\right)+2{\text{Cr}}^{3+}\left(aq\right)+6{\text{H}}_2\text{O}$
Use Table 17.1 to answer the following questions. Support your answers with calculations.
(a) Is the reaction spontaneous at standard conditions?
(b) Is the reaction spontaneous at a pH of 3.00 with all other ionic species at 0.100 M and all gases at 1.00 atm?
(c) Is the reaction spontaneous at a pH of 8.00 with all other ionic species at 0.100 M and all gases at 1.00 atm?
(d) At what pH is the reaction at equilibrium with all other ionic species at 0.100 M and all gases at 1.00 atm?

Interpretation Introduction

(a)

Interpretation:

The following redox reaction needs to be classified as spontaneous or non-spontaneous under standard conditions

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Concept introduction:

The chemical reactions that take place in an electrochemical cell are redox reactions i.e. oxidation (loss of electrons) at the anode and reduction (gain of electrons) at the cathode The standard voltage (E°) for the cell is the sum of the standard voltages of the reduction $({\text{E}}_{\text{red}}^{\text{0}})$ and oxidation $({\text{E}}_{\text{oxd}}^{\text{0}})$ potentials

$\begin{array}{l}{\text{E}}^{\text{0}}{\text{ = E}}_{\text{red}}^{\text{0}}{\text{- E}}_{\text{oxd}}^{\text{0}}\\ {\text{(or) E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}\text{ ----------(1)}\end{array}$

A given redox reaction is spontaneous if the calculated voltage is positive and non-spontaneous if the cell voltage is negative

Answer to Problem 74QAP

E° = +0.899 V. Since the value is positive, the reaction will be spontaneous under standard conditions.

Explanation of Solution

The given redox reaction is:

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Here Cr gets oxidized as the oxidation state changes from 0 to +3.

Sulfur, S gets reduced as the oxidation state changes from +6 to +4

Calculation:

The half reactions are as follows:

$\begin{array}{l}\text{Anode (oxidation): Cr(s) }\to {\text{ Cr}}^{\text{3+}}{\text{(aq) + 3e}}^{\text{-}}{\text{ E}}^0=-0.744\text{ V}\\ \text{Cathode (reduction)}:{\text{ SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 4H}}^{\text{+}}{\text{(aq) + 2e}}^{-}\to {\text{SO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}{\text{O E}}^0\text{ = +0}\text{.155 V}\\ \end{array}$

Overall reaction can be obtained by multiplying the anode reaction by 2 and cathode by 3,

${\text{The E}}^0\text{ values will remain unchanged}\text{.}$

The total number of electrons involved will be 6.

$\begin{array}{l}\text{Overall Reaction}:\\ {\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}\\ \text{Based on equation (1):}\\ {\text{E}}^{\text{0}}{\text{ = E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}}\text{ = 0}\text{.155 -(-0}\text{.744) = +0}\text{.899V}\end{array}$

Since,

${\text{E}}^0>0$

The reaction is spontaneous.

Interpretation Introduction

(b)

Interpretation:

The following redox reaction needs to be classified as spontaneous or non-spontaneous at a pH = 3.0, concentration of ionic species = 0.100 M and gases at 1.00 atm pressure.

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Concept introduction:

The cell voltage under non-standard conditions (E) is related to the standard voltage (E°) via the Nernst equation:

$\begin{array}{l}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ --------(2)}\\ \text{where n = number electrons involved in the redox reaction}\\ \text{Q = reaction quotient = }\frac{\text{[Products]}}{\text{[Reactants]}}\end{array}$

A given redox reaction is spontaneous if the calculated voltage is positive and non-spontaneous if the cell voltage is negative

Answer to Problem 74QAP

E° = +0.534 V. Since the value is positive, the reaction will be spontaneous under the given pH, concentration and pressure conditions

Explanation of Solution

The given redox reaction is:

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Based on the Nernst equation:

$\begin{array}{l}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ}\\ {\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{ln}\frac{{{\text{[Cr}}^{\text{3+}}\text{]}}^2{{\text{[P}}_{{\text{SO}}_{\text{2}}}\text{]}}^3}{{{\text{[SO}}_{\text{4}}^{\text{2-}}]}^3{{\text{[H}}^{\text{+}}\text{]}}^{\text{12}}}\end{array}$

The concentration or activity of solid and liquid species are taken equal to 1 thus,

$\begin{array}{l}\text{[Zn] = 1}\\ {\text{and [H}}_{\text{2}}\text{O] = 1}\end{array}$

Since,

$\begin{array}{l}\text{pH = 3}\text{.0}\\ \text{And,}\\ {\text{pH = -log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}={10}^{-3.0}=0.001\text{M}\end{array}$

Now,

$\begin{array}{l}\text{E°= +0}\text{.899 V}\hfill \\ \left[{\text{Cr}}^{\text{3+}}\right]{\text{ = [SO}}_{\text{4}}^{\text{2-}}]=0.1\text{00 M}\hfill \\ \begin{array}{l}{\text{P}}_{{\text{SO}}_{\text{2}}}\text{ = 1}\text{.00 atm}\\ \text{n = 6 electrons:}\end{array}\hfill \end{array}$

Putting the values,

$\begin{array}{l}\text{E = 0}\text{.899 - }\frac{\text{0}\text{.0257}}{6}\text{ln}\frac{{\text{[0}\text{.100]}}^2{\text{[1}\text{.00]}}^3}{{\text{[0}\text{.100}]}^3{\text{[0}\text{.001]}}^{\text{12}}}\\ \text{E = 0}\text{.899 - }\frac{\text{0}\text{.0257}}{6}\text{ln}\frac{1}{\text{[}{10}^{-37}\text{]}}=+0.\text{534 V}\end{array}$

Since,

${\text{ E}}^0>0$

The reaction is spontaneous.

Interpretation Introduction

(c)

Interpretation:

The following redox reaction needs to be classified as spontaneous or non-spontaneous at a pH = 8.0, concentration of ionic species = 0.100 M and gases at 1.00 atm pressure

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Concept introduction:

The cell voltage under non-standard conditions (E) is related to the standard voltage (E°) via the Nernst equation:

$\begin{array}{l}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ --------(2)}\\ \text{where n = number electrons involved in the redox reaction}\\ \text{Q = reaction quotient = }\frac{\text{[Products]}}{\text{[Reactants]}}\end{array}$

A given redox reaction is spontaneous if the calculated voltage is positive and non-spontaneous if the cell voltage is negative.

Answer to Problem 74QAP

E° = -0.058 V. Since the value is negative the reaction will be non-spontaneous under the given pH, concentration and pressure conditions

Explanation of Solution

The given redox reaction is:

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Based on the Nernst equation:

$\begin{array}{l}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ}\\ {\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{ln}\frac{{{\text{[Cr}}^{\text{3+}}\text{]}}^2{{\text{[P}}_{{\text{SO}}_{\text{2}}}\text{]}}^3}{{{\text{[SO}}_{\text{4}}^{\text{2-}}]}^3{{\text{[H}}^{\text{+}}\text{]}}^{\text{12}}}\end{array}$

Also,

$\begin{array}{l}\text{pH = 8}\text{.0}\\ \text{And,}\\ {\text{pH = -log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}{\text{] = 10}}^{\text{-pH}}={10}^{-8.0}={10}^{-8}\text{M}\end{array}$

Also,

$\begin{array}{l}\text{E°= +0}\text{.899 V}\hfill \\ \left[{\text{Cr}}^{\text{3+}}\right]{\text{ = [SO}}_{\text{4}}^{\text{2-}}]=0.1\text{00 M}\hfill \\ \begin{array}{l}{\text{P}}_{{\text{SO}}_{\text{2}}}\text{ = 1}\text{.00 atm}\\ \text{n = 6 electrons}\end{array}\hfill \end{array}$

Putting the values,

$\begin{array}{l}\text{E = 0}\text{.899 - }\frac{\text{0}\text{.0257}}{6}\text{ln}\frac{{\text{[0}\text{.100]}}^2{\text{[1}\text{.00]}}^3}{{\text{[0}\text{.100}]}^3{\text{[0}\text{.001]}}^{\text{12}}}\\ \text{E = 0}\text{.899 - }\frac{\text{0}\text{.0257}}{6}\text{ln}\frac{1}{\text{[}{10}^{-97}\text{]}}=-0.\text{058 V}\end{array}$

${\text{Since E}}^0<0.$

The reaction will be non-spontaneous.

Interpretation Introduction

(d)

Interpretation:

The pH at which the redox reaction will reach equilibrium given that the concentration of ions is 0.100 M and pressure of gasesis 1.00 atm, needs to be determined.

Concept introduction:

The cell voltage under non-standard conditions (E) is related to the standard voltage (E°) via the Nernst equation:

$\begin{array}{l}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ --------(2)}\\ \text{Here, n = number electrons involved in the redox reaction}\\ \text{Q = reaction quotient = }\frac{\text{[Products]}}{\text{[Reactants]}}\end{array}$

A given redox reaction is spontaneous if the calculated voltage is positive and non-spontaneous if the cell voltage is negative.

Answer to Problem 74QAP

The reaction reaches equilibrium at pH = 7.51

Explanation of Solution

The given redox reaction is:

${\text{3SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + 12H}}^{\text{+}}\text{(aq) + 2Cr(s) }\to {\text{3SO}}_{\text{2}}{\text{(g) + 2Cr}}^{\text{3+}}{\text{(aq) + 6H}}_{\text{2}}\text{O}$

Based on the Nernst equation:

$\begin{array}{l}{\text{E = E}}^{\text{0}}\text{ - }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ}\\ \text{At equilibrium E = 0}\text{. Therefore,}\\ {\text{E}}^{\text{0}}\text{ = }\frac{\text{0}\text{.0257}}{\text{n}}\text{lnQ}\\ {\text{ E}}^{\text{0}}\text{ = }\frac{\text{0}\text{.0257}}{\text{n}}\text{ln}\frac{{{\text{[Cr}}^{\text{3+}}\text{]}}^2{{\text{[P}}_{{\text{SO}}_{\text{2}}}\text{]}}^3}{{{\text{[SO}}_{\text{4}}^{\text{2-}}]}^3{{\text{[H}}^{\text{+}}\text{]}}^{\text{12}}}\end{array}$

Here,

$\begin{array}{l}\text{E°= +0}\text{.899 V}\hfill \\ \left[{\text{Cr}}^{\text{3+}}\right]{\text{ = [SO}}_{\text{4}}^{\text{2-}}]=0.1\text{00 M}\hfill \\ \begin{array}{l}{\text{P}}_{{\text{SO}}_{\text{2}}}\text{ = 1}\text{.00 atm}\\ \text{n = 6 electrons}\end{array}\hfill \end{array}$

Putting the values,

$\begin{array}{l}\text{0}\text{.899 = }\frac{\text{0}\text{.0257}}{6}\text{ln}\frac{{\text{[0}\text{.100]}}^2{\text{[1}\text{.00]}}^3}{{\text{[0}\text{.100}]}^3{{\text{[H}}^{\text{+}}\text{]}}^{\text{12}}}\\ \text{0}\text{.899 = }\frac{\text{0}\text{.0257}}{6}\text{ln}\frac{1}{\text{[0}\text{.100}]{{\text{[H}}^{\text{+}}\text{]}}^{\text{12}}}\\ 209.88=\text{ln}\frac{1}{\text{[0}\text{.100}]{{\text{[H}}^{\text{+}}\text{]}}^{\text{12}}}\\ {{\text{[H}}^{\text{+}}}^{\text{]}}=7.06\times {10}^{-91}\\ {\text{[H}}^{\text{+}}\text{]}=3.07\times {10}^{-8}\text{M}\\ {\text{pH = -log[H}}^{\text{+}}\text{] = -log[}3.07\times {10}^{-8}]=7.51\end{array}$