   Chapter 19.5, Problem 19.6CYU

Chapter
Section
Textbook Problem

A voltaic cell is set up with an aluminum electrode in a 0.025 M Al(NO3)3(aq) solution and an iron electrode in a 0.50 M Fe(NO3)2(aq) solution. Determine the cell potential, Ecell, at 298 K.

Interpretation Introduction

Interpretation:

The cell 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 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 :  2Al(s)  2Al3+(aq) + 6e-At cathode:Reduction : 3Fe2+(aq) +6e 3Fe(s)

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

2Al(s) + 3Fe2+(aq)  2Al3+(aq) + 3Fe(s)

Let’s calculate the Ecello of the reaction.

Ecello= ECathode0- EAnode0= - 0.45 V-(-1.66 V)=  1

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