   # Consider the following galvanic cell at 25°C: Pt | Cr 2+ (0 .30 M ),Cr 3+ (2 .0 M ) | | Co 2+ (0 .20 M ) | Co The overall reaction and equilibrium constant value are 2 Cr 2 + ( a q ) + Co 2 + ( a q ) → 2 Cr 3 + ( a q ) + Co ( s ) K = 2.79 × 10 7 Calculate the cell potential, for this galvanic cell and Δ G for the cell reaction at these conditions. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 17, Problem 87E
Textbook Problem
563 views

## Consider the following galvanic cell at 25°C: Pt | Cr 2+ (0 .30 M ),Cr 3+ (2 .0 M ) | | Co 2+ (0 .20 M ) | Co The overall reaction and equilibrium constant value are 2 Cr 2 + ( a q ) + Co 2 + ( a q ) → 2 Cr 3 + ( a q ) + Co ( s )         K =   2.79   ×   10 7 Calculate the cell potential, for this galvanic cell and ΔG for the cell reaction at these conditions.

Interpretation Introduction

Interpretation:

A reaction taking place in a galvanic cell and the equilibrium constant for the same reaction is given. The value of cell potential E and ΔG is to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

This relation is further used to determine the relation between ΔG° and K , ΔG° and E°cell .

To determine: The value of cell potential E and ΔG for the given cell reaction.

### Explanation of Solution

Given

The cell reaction is,

Pt|Cr2+(0.30M),Cr3+(2.0M)||Co2+(0.20M)|Co

The value of K is 2.79×107 .

The reaction taking place at cathode is,

Co2+(aq)+2eCo(s)E°red=0.28V

The reaction taking place at anode is,

Cr2+(aq)Cr3+(aq)+eE°ox=0.41V

Multiply oxidation half reaction with the coefficient of 2 and add both the reduction and oxidation half-reaction,

Co2+(aq)+2eCo(s)2Cr2+(aq)2Cr3++2e

The final equation is,

2Cr2+(aq)+Co2+(aq)2Cr3+(aq)+Co(s)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=0.41V+(0.28V)=0.13V

The reaction involves the transfer of 2 moles of electrons.

The cell potential is calculated using the Nernst equation,

E=E°(0.0591n)log(Q)

Where,

• E is the cell potential.
• E° is the cell potential at the standard conditions.
• Q is the activity of the species in the cell.
• n is the number of electrons involved in the reaction.

Substitute the values of Eo , n and Q in above equation as,

E=0

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