Chapter 18, Problem 18.32QP

Calculate E°, E, and ΔG for the following cell reactions.

$\begin{array}{l}(\text{a})\text{ Mg}(s)+{\text{Sn}}^{2+}(aq)\rightleftarrows {\text{Mg}}^{\text{2+}}(aq)+\text{Sn}(s)\\ [{\text{Mg}}^{2+}]=0.045M\text{, }[{\text{Sn}}^{2+}]\rightleftarrows 0.035M\\ (\text{b})\text{ 3Zn}(s)+{\text{2Cr}}^{3+}(aq)\rightleftarrows {\text{3Zn}}^{\text{2+}}(aq)+\text{2Cr}(s)\\ [{\text{Cr}}^{\text{3+}}]=0.010M,[{\text{Zn}}^{\text{2+}}]\rightleftarrows 0.0085M\end{array}$

Interpretation Introduction

Interpretation: The value of ${\text{E}}_{\text{cell}}^{\text{0}}$, $\text{E}$ and $\text{ΔG}$ of the following reactions should be calculated.

Concept introduction:

• The substance that easily be reduced in a reaction is represented as an oxidizing agent. For metal cations, a good oxidizing agent can be determined by the standard reduction potential values.
• Half-cell: In the electrochemical cell, both oxidation and reduction occurs. Oxidation occurs at the anode and reduction occurs at the cathode
• Standard electrode potential of cell is defined as the difference of reduction potential at cathode to the reduction potential at anode.
• For the cathode and anode reactions, the values are taken from the standard reduction potential table. The standard cell potential is calculated by taking the difference between the standard reduction potential values of the cathode and anode.

${\text{E}}_{\text{cell}}^{\text{0}}={\text{E}}_{\text{cathode}}^{\text{0}}-{\text{E}}_{\text{anode}}^{\text{0}}$

• ${\text{E}}_{\text{cell}}\text{=}{\text{E}}^{\text{0}}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{\left[\text{anode}\right]}{\left[\text{cathode}\right]}$
• $\begin{array}{c}\text{ΔG=}-{\text{nFE}}_{\text{cell}}^{}\\ \text{ΔG}=\text{Free}\text{energy change}\\ {\text{E}}_{\text{cell}}^{}\text{=}\text{Electrochemical}\text{cell}\text{potential}\\ \text{F}\text{=}\text{Faraday}\text{constant}\\ \text{n}\text{=}\text{number}\text{of}\text{electrons}\text{passed}\text{from}\text{anode}\text{to}\text{cathode}\end{array}$

To determine: The value of ${\text{E}}_{\text{cell}}^{\text{0}}$, $\text{E}$ and $\text{ΔG}$ of the following reactions

Answer to Problem 18.32QP

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=2.23\text{V}\\ \text{E}\text{=}2.23\text{V}\\ \text{ΔG}\text{=}-430\text{kJ/mol}\end{array}$

Explanation of Solution

Explanation

There are two half-cell reactions are involved

At anode,

${\text{Mg}}_{\left(\text{s}\right)}\underset{}{\to }{\text{Mg}}^{\text{2+}}\text{+}{\text{2e}}^{-}$

At cathode,

${\text{Sn}}^{2+}\text{+}2{\text{e}}^{-}\underset{}{\to }{\text{Sn}}_{\text{(s)}}$

Therefore,

Overall reaction can be written as,

${\text{Mg}}_{\left(\text{s}\right)}\text{+}{\text{Sn}}^{2+}\underset{}{\to }{\text{Mg}}^{\text{2+}}\text{+}{\text{Sn}}_{\left(\text{s}\right)}$

${\text{E}}_{\text{cell}}^{\text{0}}$, $\text{E}$ and $\text{ΔG}$ is calculated for the above reaction.

Given,

From standard reduction potential table,

${\text{E}}_{\text{anode}}^{\text{0}}=-2.37\text{V}$

${\text{E}}_{\text{cathode}}^{\text{0}}=-0.14\text{V}$

$\begin{array}{c}\\ \left[{\text{Mg}}^{2+}\right]=0.045\\ \left[{\text{Sn}}^{2+}\right]=0.035\end{array}$

${\text{E}}_{\text{cell}}^{\text{0}}$ can be calculated by using the given equation,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}={\text{E}}_{\text{cathode}}^{\text{0}}-{\text{E}}_{\text{anode}}^{\text{0}}\\ ={\text{E}}_{{\text{Sn}}^{\text{2+}}\text{/Sn}}^{\text{0}}-{\text{E}}_{{\text{Mg}}^{\text{+}}\text{/Mg}}^{\text{0}}\\ =-0.14\text{V}-\left(-2.37\right)\text{V}\\ =2.23\text{V}\end{array}$

$\text{E}$ can be calculated by using the given equation,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}^{\text{0}}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{\left[\text{anode}\right]}{\left[\text{cathode}\right]}$

Therefore,

$\begin{array}{c}{\text{E}}_{\text{cell}}={\text{E}}^{\text{0}}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{\left[{\text{Mg}}^{2+}\right]}{\left[{\text{Sn}}^{2+}\right]}\\ =2.23\text{V}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{0.045}{0.035}\\ =2.23\text{V}\end{array}$

${\text{ΔG}}^0$ can be calculated by using the given equation,

$\begin{array}{c}{\text{ΔG}}^0\text{=}-{\text{nFE}}_{\text{cell}}^{\text{0}}\\ {\text{ΔG}}^0=\text{Standard}\text{free}\text{energy change}\\ {\text{E}}_{\text{cell}}^{\text{0}}\text{=}\text{Standard}\text{electrochemical}\text{cell}\text{potential}\\ \text{F}\text{=}\text{Faraday}\text{constant}\\ \text{n}\text{=}\text{number}\text{of}\text{electrons}\text{passed}\text{from}\text{anode}\text{to}\text{cathode}\end{array}$

Therefore,

$\begin{array}{c}{\text{ΔG}}^0\text{=}-{\text{nFE}}_{\text{cell}}\\ \text{=}-({\text{2e}}^{-}\text{×}\text{96500}\text{J/V}\text{mol}\text{×}\text{2}\text{.23}\text{V)}\\ \text{=}-430\text{kJ/mol}\end{array}$

Interpretation Introduction

Interpretation: The value of ${\text{E}}_{\text{cell}}^{\text{0}}$, $\text{E}$ and $\text{ΔG}$ of the following reactions should be calculated.

Concept introduction:

• The substance that easily be reduced in a reaction is represented as an oxidizing agent. For metal cations, a good oxidizing agent can be determined by the standard reduction potential values.
• Half-cell: In the electrochemical cell, both oxidation and reduction occurs. Oxidation occurs at the anode and reduction occurs at the cathode
• Standard electrode potential of cell is defined as the difference of reduction potential at cathode to the reduction potential at anode.
• For the cathode and anode reactions, the values are taken from the standard reduction potential table. The standard cell potential is calculated by taking the difference between the standard reduction potential values of the cathode and anode.

${\text{E}}_{\text{cell}}^{\text{0}}={\text{E}}_{\text{cathode}}^{\text{0}}-{\text{E}}_{\text{anode}}^{\text{0}}$

• ${\text{E}}_{\text{cell}}\text{=}{\text{E}}^{\text{0}}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{\left[\text{anode}\right]}{\left[\text{cathode}\right]}$
• $\begin{array}{c}\text{ΔG=}-{\text{nFE}}_{\text{cell}}^{}\\ \text{ΔG}=\text{Free}\text{energy change}\\ {\text{E}}_{\text{cell}}^{}\text{=}\text{Electrochemical}\text{cell}\text{potential}\\ \text{F}\text{=}\text{Faraday}\text{constant}\\ \text{n}\text{=}\text{number}\text{of}\text{electrons}\text{passed}\text{from}\text{anode}\text{to}\text{cathode}\end{array}$

To determine: The value of ${\text{E}}_{\text{cell}}^{\text{0}}$, $\text{E}$ and $\text{ΔG}$ of the following reactions

Answer to Problem 18.32QP

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}=0.02\text{V}\\ \text{E}\text{=}0.04\text{V}\\ \text{ΔG}\text{=}-23\text{kJ/mol}\end{array}$

Explanation of Solution

Explanation

There are two half-cell reactions are involved

At anode,

$\text{3}\left[{\text{Zn}}_{\left(\text{s}\right)}\underset{}{\to }{\text{Zn}}^{\text{2+}}\text{+}{\text{2e}}^{-}\right]$

At cathode,

$\text{2}\left[{\text{Cr}}^{\text{3+}}\text{+}{\text{3e}}^{-}\underset{}{\to }{\text{Cr}}_{\left(\text{s}\right)}\right]$

Therefore,

Overall reaction can be written as,

${\text{3Zn}}_{\left(\text{s}\right)}\text{+}2{\text{Cr}}^{3+}\underset{}{\to }3{\text{Cr}}^{\text{2+}}\text{+}{\text{2Cr}}_{\left(\text{s}\right)}$

${\text{E}}_{\text{cell}}^{\text{0}}$, $\text{E}$ and $\text{ΔG}$ is calculated for the above reaction.

Given,

From standard reduction potential table,

${\text{E}}_{\text{anode}}^{\text{0}}=-0.76\text{V}$

${\text{E}}_{\text{cathode}}^{\text{0}}=-0.74\text{V}$

$\begin{array}{c}\\ \left[{\text{Zn}}^{2+}\right]=0.0085\\ \left[{\text{Cr}}^{3+}\right]=0.010\end{array}$

${\text{E}}_{\text{cell}}^{\text{0}}$ can be calculated by using the given equation,

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{0}}={\text{E}}_{\text{cathode}}^{\text{0}}-{\text{E}}_{\text{anode}}^{\text{0}}\\ ={\text{E}}_{{\text{Cr}}^{\text{3+}}\text{/Cr}}^{\text{0}}-{\text{E}}_{{\text{Zn}}^{\text{2+}}\text{/Zn}}^{\text{0}}\\ =-0.74\text{V}-\left(-0.76\right)\text{V}\\ =0.02\text{V}\end{array}$

$\text{E}$ can be calculated by using the given equation,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}^{\text{0}}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{\left[\text{anode}\right]}{\left[\text{cathode}\right]}$

Therefore,

$\begin{array}{c}{\text{E}}_{\text{cell}}={\text{E}}^{\text{0}}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{{\left[{\text{Zn}}^{2+}\right]}^3}{{\left[{\text{Cr}}^{3+}\right]}^2}\\ =0.02\text{V}-\frac{\text{0}\text{.0592}\text{V}}{\text{n}}\log \frac{{\left(0.0085\right)}^3}{{\left(0.010\right)}^2}\\ =0.04\text{V}\end{array}$

${\text{ΔG}}^0$ can be calculated by using the given equation,

$\begin{array}{c}{\text{ΔG}}^0\text{=}-{\text{nFE}}_{\text{cell}}^{\text{0}}\\ {\text{ΔG}}^0=\text{Standard}\text{free}\text{energy change}\\ {\text{E}}_{\text{cell}}^{\text{0}}\text{=}\text{Standard}\text{electrochemical}\text{cell}\text{potential}\\ \text{F}\text{=}\text{Faraday}\text{constant}\\ \text{n}\text{=}\text{number}\text{of}\text{electrons}\text{passed}\text{from}\text{anode}\text{to}\text{cathode}\end{array}$

Therefore,

$\begin{array}{c}{\text{ΔG}}^0\text{=}-{\text{nFE}}_{\text{cell}}\\ \text{=}-({\text{6e}}^{-}\text{×}\text{96500}\text{J/V}\text{mol}\text{×}\text{0}\text{.04}\text{V)}\\ \text{=}-23\text{kJ/mol}\end{array}$