Question

An electrochemical cell is based on the following two half-reactions:
Ox: Pb(s)→Pb2+(aq, 0.15 M )+2e−
Red:MnO−4(aq, 1.70 M )+4H+(aq, 1.5 M )+3e−→ MnO2(s)+2H2O(l)

Compute the cell potential at 25 ∘C.

Answer 1

Given that

Red: MnO4−(aq, 1.70 M )+4H+(aq, 1.5 M ) + 3 e− → MnO2(s)+2H2O(l) + 1.70 ------------(1)

Ox: Pb(s) → Pb2+(aq, 0.15 M ) + 2e− + 0.126 -------------- (2)

Multiply (1) with 2 and (2) with 3,

2MnO4−(aq)+ 8H+(aq) + 6 e− → 2MnO2(s)+ 4 H2O(l)

3Pb(s) → 3Pb2+(aq) + 6e−

--------------------------------------------------------------------------------------------

2MnO₄−(aq)+ 8H+(aq) +   3Pb(s) ----------->  MnO2(s)+2H2O(l) +  3Pb²⁺(aq)

E^o_cell = E^o_reduction of reaction at cathode+E^o_oxidation of reaction at anode

= 1.70 + 0.126 V

= +1.826

E^o_cell = +1.826

no of electrons transferred n = 6

Cell potential:

2MnO₄−(aq)+ 8H+(aq) +   3Pb(s) ----------->  MnO2(s)+2H2O(l) +  3Pb²⁺(aq)

E^o_cell = +1.826

no of electrons transferred n = 6

[Pb²⁺] =  0.15 M

[MnO₄-] = 1.7 M

[H⁺] = 1.5 M

E_cell = E^o_cell - RT/nF In {[Pb²⁺]³/ [MnO₄-]² [H⁺]⁸} (solids cannot be taken into consideration)

=  E^o_cell - (0.059/n) Iog {[Pb²⁺]³/ [MnO₄-]² [H⁺]⁸}

= 1.826 - (0.059/6) log  {[0.15]³/ [1.7]² [1.5]⁸}

= +1.87 V

E_cell =  +1.87 V

Therefore, cell potential = +1.87 V