   Chapter 19, Problem 31PS

Chapter
Section
Textbook Problem

One half-cell in a voltaic cell is constructed from a silver wire electrode in a 0.25 M solution of AgNO3. The other half-cell consists of a zinc electrode in a 0.010 M solution of Zn(NO3)2. Calculate the cell potential.

Interpretation Introduction

Interpretation:

The potential for the given voltaic cell has to be determined.

Concept introduction:

Electrochemical cells under nonstandard conditions:

Basically standard electrode potentials are determined at

1.00 M solution concentration

1.00 atm for gases

Pure liquids or solids

At 250C

Under the non –standard condition electrode potential can be calculated by using the Nernst equation.

According to the Nernst equation, the cell potentials are related to concentrations of reactants and products and to temperature as follows.

E = E0-(RTnF)lnQ

Let’s write an each variable in the Nernst equation.

R= Gas constant = 8.314j/k.molT= Temperature (K)n = number of moles of  transferred in between oxidizing and reducing agentsF = Faraday constant = 9.6648533289 × 104C/mol

At 298K the Nernst equation is as follows

E = E0 - 0.0257n lnQ at 298 K

Explanation

Let’s write an each half cell reaction.

At anode:Oxiation : Zn(s)  Zn2+(aq) + 2e-(aq)At cathode:Reduction : 2Ag(s) + 2e2Ag+(aq)

The overall reaction:

Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)

Let’s calculate the Ecello value of the reaction.

Ecello= ECathode0- EAnode0= 0.799  V-(-0.763V)= 1.562 V

Calculate the potential of the cell by using Nernst equation:

E = E0 - 0

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