Question

79. Two half-cells, PtlFe3+(aq, 0.50 M), Fe2 +(aq, 1.0 x 10-5 M) and Hg (aq, 0.020 M)IHg, are constructed and then linked together to form a voltaic cell. Which electrode is the anode? What will be the potential of the voltaic cell at 298 K:

Answer 1

i)

Hg/Hg²⁺(aq,0.020M) is anode

ii)

E°_cell = E°_red,cathode- E°_red,anode

= + 0.770V - (+0.850V)

= - 0.080V

iii)

Oxidation half reaction

Hg(s) -------> Hg²⁺(aq) + 2e

Reduction half reaction

2Fe³⁺(aq) + 2e --------> 2Fe²⁺(aq)    

Overall reaction

Hg(s) + 2Fe³⁺(aq) -------> 2Fe²⁺(aq) + Hg²⁺(aq)

Q= [Hg²⁺][Fe²⁺]²/[Fe³⁺]²

  Q =( (0.020M)(0.00001M)²)/(0.50M)²

Q= 8.0×10^-12

^{Number of electron transfer,n = 2}

^iv)

^{Nernst equation at 298.15K is as follows}

E_cell = E°_cell - (0.0592V/n)logQ

= -0.080V - (0.0592V/2)log(8.0×10^-12)

= -0.080V + 0.328V

= 0.248V