Question

There is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+.

A. Write an equation for the half-reaction occurring at each electrode. Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.
B. What minimum voltage is necessary to drive the reaction? Vmin =

Answer 1

-1.05 V

Step-by-step explanation:

At the anode;

Sn(s) -----> Sn^2+(aq) + 2e

At the cathode;

Mn^2+(aq) + 2e ------> Mn(s)

Given that it is an electrolytic cell,

E°cell= E°cathode - E°anode

But;

E°cathode= -1.19 V

E°anode= -0.14 V

E°cell= -1.19-(-0.14)

E°cell= -1.05 V

Therefore

Vmin = -1.05 V