Question

Cathodic Hydrogen Formation

1. The delta Go for the following reaction is -0.28eV, yet the reaction does not occur at a measurable rate. However, if one inserts a Sn wire into an electrochemical cell and applies a cathotic potential of -0.3V, hydrogen gas bubbles off the surface of the wire. Explain. Sn + 2H+ --> Sn2+ + H2(g)

Answer 1

Given reaction

Sn+ 2H⁺ -- > H2 (g) + Sn²⁺

Oxidation half:

Sn --- > Sn²⁺ + 2e^-                  E_red = -0.14 V

Reduction half

2H⁺+ 2e^- -- > H₂(g )                E_red = 0.0 V

The overall reaction shows Sn oxidized and H+ is reduced to H₂ g. This is done when H+ accepts the electrons from Sn. If we look in the electrochemical series then we can overseer that there is not much difference in reduction potential between two electrodes.

Due to less difference, reaction rate is less. When we apply cathodic current to Sn wire then there is flow of electrons from Sn wire to solution and electrons on Sn are taken by H⁺ ions. As we know H+ take electrons and form hydrogen gas.

Sot the rate increase.