Question

Derive a balanced equation for the reaction occurring in the cell:
Fe(s)|Fe²⁺(aq)||Fe³⁺(aq),Fe²⁺(aq)|Pt(s)

a.) If E?cell = 1.21 V, calculate ?G? for the reaction.

b.) If E?cell=1.21V, calculate the equilibrium constant for the reaction.

c.) Use the Nernst equation to determine the potential for the cell:
Fe(s)|Fe²⁺(aq,1.0�10^-3M)||Fe³⁺(aq,1.0�10^-3M),Fe²⁺(aq,0.10M)|Pt(s)

Answer 1

LHS Fe(S) Fe²⁺ + 2e (Oxidation)

RHS 2(Fe³⁺ + e Fe²⁺ ) (Reduction)

To balance charge multiply the reduction reaction by 2 and the add the two equations

The balanced equation for the cell reaction is

Fe(s) + 2 Fe³⁺3 Fe⁺⁺    

a) ? G⁰ = - nFE

n, the number of electrons involved in the redox reaction = 2

F is the Faraday = 96500 Coulombs

V = 1.21 V

? G = - 2 x 96500 Coulombs x 1.21 Volt= 233,530 volt coulombs

1 volt coulomb = 1 Joule

Therefore ?G = - 233,530 J = - 233.530 kJ

b) - ? G⁰ = RT ln K

ln K = - ? G⁰/ RT

ln K = 2.303 log K

Therefore

log K = - ?G⁰/2.303RT

R , the gas constant = 8.314J K^-1mol^-1

T ,the temperatue is not given .Let us assume a value of 298K

log K = -(- 233,530)J/ 2.303 x 8.314 J K^-1mol^-1 x 298 K

         = 40.9281

K = anti log of 40.9281

K = 8.47 x 10⁴⁰

^C) LHs Fe(s) = Fe⁺⁺ (1x 10^-3M) +2e

RHS [Fe⁺⁺⁺ ( 1x10^-3) + e = Fe ⁺⁺ (0.1M)}2

Adding Fe(s) +2Fe(1x10^-3) = Fe⁺⁺(1x10^-3) + 2Fe⁺⁺(0.1M)

NERNST EQUATION

E = E⁰- RT/nF ln [Fe++ 1x10^-3][Fe++ 0.1]²/[Fe+++ 1x10^-3]²

E = E⁰ - 0.0591/2 log [1x 10^-3] [0.1]² / [1x10^-3]² at 298 K

= 1.21 -0.02955x log10

=1.21 -0.02955x1

=1.043 V

=1,23955 V