Question

Consider a galvanic cell that contains a Zn metal electrode and a 0.10 M Zn(NO3)2 solution in one half-cell and a Sn metal electrode and a 0.10 M Sn(NO3)2 solution in the other half-cell. If the measured Ecell value is +0.60 V and the Zn2+/Zn reduction potential is assumed to be -0.79 V, what is the Sn2+/Sn reduction potential? Show all work

Answer 1

To solve this question we need to have basic knowledge of reaction taking place.

reaction taking place is:

Zn+ Sn²⁺ --->Zn²⁺ + sn

Here Zn2+/Zn is anode and Sn2+/Sn is cathode

E cell = Eo cell - (R*T/nF) ln {[Sn2+]/[Zn2+]}

[Zn2+] = [Zn(NO3)2] = 0.1M

[Sn2+] = [Sn(NO3)2] = 0.1M

E cell = Eo cell - (0.059/2) ln {[Sn2+]/[Zn2+]}

Because: RT/F turns out to be 0.059 V

n=number of electrons being transferred = 2

putting values:

Ecell= Eo cell - (0.059/2) ln {0.1/0.1]}

so,

Ecell =Eo cell

Ecell = Eo cathode - Eo anode

0.6 = Eo cathode - (-0.79)

Eo cathode = -0.19 V

so, Sn2+/Sn reduction potential = -0.19 V