Question

During the electrolysis of MnCl2(aq), what is the product formed at the cathode? Take into account an overvoltage of 0.57 V for the production of O2(g), 0.45 V for the production of H2(g) and 0.30 V for the production of Cl2(g). A) Cl2(g) B) H2O(l ) C) H2(g ) D) Mn(s)

Answer 1

Reduction half reaction occurs to cathode during electrolysis.

The possible reduction reactions during the electrolysis of MnCl₂(aq) are

Mn²⁺(aq) +2e^-Mn(s) Eo(Mn²⁺/Mn) = -1.70 V

2H⁺ (aq) + 2e^- H₂(g) Theoretical Eo (H⁺/H2) = 0.00 V

  Practical   Eo (H⁺/H2) = 0.00 V+ overpotential

  =0.00 V+0.45 V

=0.45V

The species with more positive reduction potential value undergoes reduction more preferably.

On comparing the reduction potential values, we can conclude that H+ will undergo reduction preferably at cathode and the product at cathode will be  H₂(g) .

Option (c)  H₂(g) is correct option.