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.

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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 Δ G 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 E cell is calculated using Nernst formula, E = E ° − ( R T n F ) ln ( Q ) At room temperature the above equation is specifies as, E = E ° − ( 0.0591 n ) log ( Q ) This relation is further used to determine the relation between Δ G ° and K , Δ G ° and E ° cell . To determine: The value of cell potential E and Δ G for the given cell reaction. Explanation Given The cell reaction is, Pt | Cr 2 + ( 0.30 M ) , Cr 3 + ( 2.0 M ) | | Co 2 + ( 0.20 M ) | Co The value of K is 2.79 × 10 7 . The reaction taking place at cathode is, Co 2 + ( a q ) + 2 e − → Co ( s ) E ° red = − 0.28 V The reaction taking place at anode is, Cr 2 + ( a q ) → Cr 3 + ( a q ) + e − E ° ox = 0.41 V Multiply oxidation half reaction with the coefficient of 2 and add both the reduction and oxidation half-reaction, Co 2 + ( a q ) + 2 e − → Co ( s ) 2Cr 2 + ( a q ) → 2 Cr 3 + + 2 e − The final equation is, 2Cr 2 + ( a q ) + Co 2 + ( a q ) → 2 Cr 3 + ( a q ) + Co ( s ) The overall cell potential is calculated as, E ° cell = E ° ox + E ° red = 0.41 V+ ( − 0.28 V ) = 0.13 V The reaction involves the transfer of 2 moles of electrons. The cell potential is calculated using the Nernst equation, E = E ° − ( 0.0591 n ) 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 E o , n and Q in above equation as, E = 0

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Publisher: Cengage Learning

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

The value of K is 2.79×107 .

The reaction taking place at cathode is,

Co2+(aq)+2e−→Co(s) E°red=−0.28 V

The reaction taking place at anode is,

Cr2+(aq)→Cr3+(aq)+e− E°ox=0.41 V

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

Co2+(aq)+2e−→Co(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=0.41 V+ (−0.28 V)=0.13 V

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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