   Chapter 19.5, Problem 19.7CYU

Chapter
Section
Textbook Problem

The half-cells Ag+(aq. 1.0 M)|Ag(s) and H+(aq, ? M)|H2(1.0 bar) are linked by a salt bridge to create a voltaic cell. With the silver electrode as the cathode, a value of 0.902 V is recorded tor kcell at 298 K. Determine the concentration of H+ and the pH of the solution.

Interpretation Introduction

Interpretation:

The concentration of H+ and pH of 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 : H2(g)  2H+(aq) + 2e-At cathode:Reduction : 2Ag+(aq) +2 e 2Ag(s)

By adding these two half cell reactions we get an overall reaction:

H2(g) + 2Ag+(aq)  2Ag(s) + 2H+(aq)

Let’s calculate the Ecello of the reaction:

Ecello= Ecathodeo-EAnodeo= 0.80 V - 0.00 V= 0.80 V

The potential of the reaction is 0.902 V

Calculate the pH of the cell by using Nernst equation:

E = E0 - 0.0257n lnQ at 298 K

Substitute the given values in the above equation

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