Electrochemistry

Question

What is the potential of a cell made up of Zn/Zn2+ and Cu2+/Cu, half cell at 76°c. If [Zn2+]=0.25M and [Cu2+]=0.15M

Given that Zn/Zn2+ E°=-0.76V, Cu2+/Cu E°=0.34

Answer 1

Using standard reduction potentials:

Zn2+ + 2e- ==> Zn Eº = -0.76 V

Cu2+ + 2e- ==> Cu Eº = +0.34 V

Overall redox:

Cu2+ + Zn ==> Cu(s) + Zn2+

Cathode = Cu

Anode = Zn

Eºcell = cathode - anode = 0.34 - (-0.76) = 1.1 V

Using the Nernst equation to find Ecell under the non-standard conditions, we have ...

Ecell = Eº - RT/nF ln Q

Eº = 1.1

R = 8.314

T = 76ºC + 273 = 349

n = 2

F = 96,500

Q = [Zn2+] / [Cu2+] = 0.25 / 0.15 = 1.67

Ecell = 1.10 - (8.314)(349) / (2)(96500) ln 1.67

Ecell = 1.10 - (0.015)(0.51)

Ecell = 1.10 - 0.0077

Ecell = 1.09 V

Be sure to check the math