Chapter 18.4, Problem 18.4WE

Interpretation Introduction

Interpretation:

The standard free energy change of the given reaction 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.

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

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

Answer to Problem 18.4WE

The standard free energy change is $2.53\times {10}^{3}kJ/mol$

Explanation of Solution

There are two half-cell reactions involved

At anode,

$2A{u}_{\left(s\right)}\underset{}{\to }2A{u}_{\left(aq\right)}^{3+}+6e^{-}$${\text{E}}_{\text{anode}}^{\text{0}}=1.50\text{V}$

At cathode,

$3C{a}_{\left(aq\right)}^{2+}+6e^{-}\stackrel{}{\to }3C{a}_{\left(s\right)}$${\text{E}}_{\text{cathode}}^{\text{0}}=-2.87\text{V}$

${\text{ΔG}}^0$ is calculated for the above reaction.

Given,

From standard reduction potential table,

${\text{E}}_{\text{anode}}^{\text{0}}={\text{E}}_{{\text{Au}}^{\text{3+}}\text{/Au}}^{\text{0}}=1.50\text{V}$

${\text{E}}_{\text{cathode}}^{\text{0}}={\text{E}}_{{\text{Ca}}^{\text{2+}}\text{/Ca}}^{\text{0}}=-2.87\text{V}$

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

$\begin{array}{c}{E}_{cell}^0=E_{cathode}^0-E_{anode}^0\\ =E_{C{a}^{2+}/Ca}^0-E_{A{u}^{3+}/Au}^0\\ =-2.87V-1.50V\\ =-4.37V\end{array}$

${\text{ΔG}}^0$ 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,

$\begin{array}{c}\Delta G^0=-\left(6mol{e}^{-}\right)\times \left(96500J/Vmol{e}^{-}\right)\times \left(-4.37V\right)\\ =2.53\times {10}^{6}J/mol\\ =2.53\times {10}^{3}kJ/mol\end{array}$

The standard free energy change is $2.53\times {10}^{3}kJ/mol$

Conclusion

The standard free energy change was calculated as $2.53\times {10}^{3}kJ/mol$