Question

In: Chemistry

For the voltaic cell, Cr(s) │ Cr3+ (aq, 0.24 M) ││ Fe2+ (aq, (?) M) │...

For the voltaic cell, Cr(s) │ Cr3+ (aq, 0.24 M) ││ Fe2+ (aq, (?) M) │ Fe(s) , Ecell is 0.33 V. Calculate the concentration of Fe2+ (M). Reduction potential for Cr3+(aq)/Cr(s) is -0.74 V, Fe2+(aq)/Fe(s) is -0.44 V. Enter number to 2 decimal places.

Solutions

Expert Solution

Oxidation half reaction

2Cr(s) ---------> 2Cr3+(aq) + 6e-red = -0.74V

Reduction half reaction

3Fe2+(aq) + 6e- -------> 3Fe(s) E°red = -0.44V

number of electron transfer , n = 6

cell = E°red, cathode - E°red, anode

cell = -0.44V - (- 0.74V)

cell = 0.30V

overall reaction is

3Fe2+(aq) + 2Cr(s) -------> 3Fe(s) + 2Cr3+(aq)

reaction quotient , Q = [Cr3+]2/[Fe2+]3

Q = (0.24)2/ [Fe2+]3 = 0.0576/[Fe2+]3

Nernst equation at 25℃ is

Ecell = E°cell - (0.0592/n)logQ

0.33V = 0.30V - (0.0592V/6)log( 0.0576/[Fe2+]3)

(0.0592V/6)log(0.0576/[Fe2+]3) = -0.03V

log(0.0576/[Fe2+]3) = - 3.041

( 0.0576/[Fe2+]3 ) = 0.0009099

[Fe2+]3 = 0.0576/ 0.0009099

[Fe2+]3 = 63.30

[Fe2+] = 3.99M


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