Chapter 22, Problem 58AP

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction is to be calculated with the standard reduction potential.

Concept introduction:

The change in free energy is called Gibbs free energy and is denoted as $\Delta G^o$.

If $\Delta G^o$ is negative, it indicates that the electrical work has been done by the system and if it is positive then it means the electrical work has been done on the system.

The standard EMF of a cell is represented as follows:

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

The standard Gibbs free energy change is calculated by the expression as:

$\Delta G^o=-n\text{F}{E}_{cell}^o$

Here, $n$ is the moles of electrons are transferred per mole of reaction,$\Delta G^o$ is the standard Gibbs free energy change, $E_{cell}^o$ is the standard cell potential, and $\text{F}$ is the Faraday’s constant.

The equilibrium constant is calculated by the expression as:

$E_{cell}^o=\frac{0.0592}{n}\log \left(\text{K}\right)$

The relation between free energy change and standard free energy change is as:

$\Delta G=\Delta G^{\text{o}}+\text{R}T\ln Q$

Here, $\Delta G$ is free energy change, $\Delta G^o$ is standard free energy change, $Q$ is the reaction quotient, $\text{R}$ is the gas constant, and $T$ is the temperature.

At equilibrium the above equation is reducedto the expression:$\Delta G^{\text{o}}\text{=}-\text{R }T\text{lnK }$