Given the following two standard reduction potentials, solve for the standard reduction potential of the half-reaction M 3 + + e − → M 2 + ( Hint : You must use the extensive property ∆ G ° to determine the standard reduction potential.)

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Interpretation Introduction Interpretation: The standard reduction potential of two half-cells is given. The standard reduction potential for the given half-reaction is to be calculated. Concept introduction: The standard reduction potential is the potential of the cell at the standard conditions like 1.0 M concentration and a pressure of 1 atm at the given temperature conditions. To determine: The standard reduction potential of the given half-cell reaction. Explanation The given reactions are as follows, M 3 + + 3 e − → M E ° = − 0.10 V (1) M 2 + + 2 e − → M E ° = − 0.50 V (2) Subtract both the equations to obtain the given reaction as, M 3 + + e − → M 2 + (3) The cell potential for the equation ( 3 ) is calculated using the relationship between cell potential and change in Gibbs free energy, Δ G ° = − n F E ° . Where the following terms are defined as follows, Δ G ° is the change in Gibbs free energy. n is the number of electrons involved in the cell reaction. F is the Faraday’s constant ( 96485 C / mole e − ) . E ° is the cell potential. The change in Gibbs free energy for equation ( 1 ) is calculated as follows, ΔG° 1 =-nFE° 1 =- ( 3 ) ×96485 × ( -0 .10 V ) =28,945

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Interpretation Introduction

Interpretation:

The standard reduction potential of two half-cells is given. The standard reduction potential for the given half-reaction is to be calculated.

Concept introduction:

The standard reduction potential is the potential of the cell at the standard conditions like 1.0 M concentration and a pressure of 1 atm at the given temperature conditions.

To determine: The standard reduction potential of the given half-cell reaction.

Explanation of Solution

The given reactions are as follows,

M3++3e−→M E°=−0.10V (1)

M2++2e−→M E°=−0.50 V (2)

Subtract both the equations to obtain the given reaction as,

M3++e−→M2+ (3)

The cell potential for the equation (3) is calculated using the relationship between cell potential and change in Gibbs free energy, ΔG°=−nFE° .

Where the following terms are defined as follows,

ΔG° is the change in Gibbs free energy.

n is the number of electrons involved in the cell reaction.

F is the Faraday’s constant (96485 C/mole e−) .

E° is the cell potential.

The change in Gibbs free energy for equation (1) is calculated as follows,

ΔG°1=-nFE°1=-(3)×96485 ×(-0.10 V)=28,945

Given the following two standard reduction potentials, solve for the standard reduction potential of the half-reaction M 3 + + e − → M 2 + ( Hint : You must use the extensive property ∆ G ° to determine the standard reduction potential.)

Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach