Question

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.

Answer 1

Oxidation half reaction

2Cr(s) ---------> 2Cr³⁺(aq) + 6e^- E°_red = -0.74V

Reduction half reaction

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

number of electron transfer , n = 6

E°_cell = E°_{red, cathode} - E°_{red, anode}

E°_cell = -0.44V - (- 0.74V)

E°_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

E_cell = 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