Question

An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.29 M )+2e− Red: MnO−4(aq, 1.30 M )+4H+(aq, 1.7 M )+3e−→ MnO2(s)+2H2O(l) show work

Answer 1

Consider reaction taking place in the cell.

At anode: Pb (s) Pb ²⁺ (aq) + 2 e^-

At cathode : MnO ₄^-(aq) + 4 H ⁺ + 3 e^- MnO₂(s) + 2 H₂O (l)

Overall reaction : Pb (s) + MnO ₄^-(aq) + 4 H ⁺ + 3 e^- Pb ²⁺ (aq) + 2 e^- +  MnO₂(s) + 2 H₂O (l)

Balance reaction for electrons, we get

3 Pb (s) +2 MnO ₄^-(aq) + 8 H ⁺ 3 Pb ²⁺ (aq) +2 MnO₂(s) + 4 H₂O (l)

Here no of electrons transferred = 6

Standard emf of the cell is calculated as

E ⁰ cell = E ⁰ cathode - E ⁰ anode  

= 1.692 - ( -0.126 )

= 1.692+0.126

= 1.818 V

Emf of cell is calculated by using Nerns't equation as

E cell = E ⁰ cell - 2.303 R T / n F log [ Pb ²⁺ ] ³ / [MnO ₄^-  ] ² [H ⁺ ] ⁸

We have 2.303 R T / F = 0.0591 at 25 ⁰ C then

E cell = E ⁰ cell - 0.0591 / n log [ Pb ²⁺ ] ³ / [MnO ₄^-  ] ² [H ⁺ ] ⁸

= 1.818 - 0.0591/ 6 log ( 0.29) ³ / (1.30) ² ( 1.70) ⁸

= 1.818 - 0.00985 log 2.069 x 10 ^-04

= 1.818 - 0.00985 x ( - 3.684)

= 1.818 + 0.0363

= 1.854 V