   # Consider the cell described below: Zn | Zn 2+ ( 1.00 M ) | | Cu 2+ ( l .00 M ) | Cu Calculate the cell potential after the reaction has operated long enough for the [Zn 2+ ] to have changed by 0.20 mol/L. (Assume T =25°C.) ### 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 73E
Textbook Problem
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## Consider the cell described below: Zn | Zn 2+ ( 1.00   M ) | | Cu 2+ ( l .00 M ) | Cu Calculate the cell potential after the reaction has operated long enough for the [Zn2+] to have changed by 0.20 mol/L. (Assume T=25°C.)

Interpretation Introduction

Interpretation:

The standard line notation for a cell is given. The value of cell potential after the reaction has operated long enough for the [Zn2+] to have changed by 0.20mol/L 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 Nernst formula is defined as follows,

E=E°(RTnF)ln(Q)

To determine: The value of cell potential after the reaction has operated long enough for the [Zn2+] to have changed by 0.20mol/L .

### Explanation of Solution

Given

The standard line notation is,

Zn|Zn2+(1.00M)||Cu2+(1.00M)|Cu

The reaction taking place at cathode is,

Cu2++2eCuE°red=0.34V

The reaction taking place at anode is,

ZnZn2++2eE°ox=0.76V

Add both the oxidation and reduction half reaction,

Cu2++2eCuZnZn2++2e

The final equation is,

Cu2++ZnCu+Zn2+

The value of E°cell is calculated as,

E°cell=E°ox+E°red=0.76V+0.34V=1

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