   Chapter 18, Problem 136CWP

Chapter
Section
Textbook Problem

# Consider a galvanic cell based on the following theoretical half-reactions: What is the value of ∆G° and K for this cell?

Interpretation Introduction

Interpretation:

The value of ΔG° and K for the given cell is to be calculated.

Concept introduction:

The species with higher oxidation potential undergoes reduction reaction at the cathode, while the species with lower reduction potential value undergoes oxidation at the anode.

The relationship between Gibbs free energy change and cell potential is given by the formula,

ΔG°=nFE°cell

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

ΔG°=RTlnK

To determine:

The value of ΔG° and K for the given cell is to be calculated.

The value of ΔG° is -312.611kJ_ .

Explanation

The reaction taking place at the cathode is,

M4++4eME°red=0.66V

The reaction taking place at the anode is,

NN3++3eE°OX=0.39V

Multiply reduction half-reaction with a coefficient of 3 and oxidation half-reaction with a coefficient of 4 and then add both the reactions as,

3M4++12e3M4N4N3++12e

The final equation is,

3M4++4N3M+4N3+

The value of E°cell is given as,

E°cell=E°ox+E°red

Where,

• E°ox is the oxidation potential of the electrode.
• E°red is the reduction potential of the electrode.

Substitute the values of E°ox and E°red in the above equation as,

E°cell=E°ox+E°red=0.39V+0.66V=0.27V

The reaction involves the transfer of 12 moles of electrons.

The relationship between Gibbs free energy change and cell potential is given by the formula,

ΔG°=nFE°cell

Where,

• ΔG° is the Gibbs free energy change at the standard conditions.
• n is the number of moles of electrons that are involved in the reaction.
• F is the Faraday’s constant (96,485C/molee)
• E° is the cell potential at the standard conditions.

Substitute the value of n , F and E° in the above equation as,

ΔG°=nFE°cell=12e×96485Cmolee×0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 