What is the value of E for a cell in which [Cu2+] = 1.0 x 10-3 M and [Zn2+] = 1.0 M?

Question

The value of Eºcell for the reaction Zn(s) + Cu2+(aq) <=> Zn2+(aq) + Cu(s) is 1.10 V. What is the value of E for a cell in which [Cu2+] = 1.0 x 10-3 M and [Zn2+] = 1.0 M?

The correct answer is 1.01

Answer 1

Zn(s) + Cu²⁺(aq) ==> Zn²⁺(aq) + Cu(s) Eº = 1.10 V

To find E_cell under non standard conditions, we use the Nernst equation

E_cell = Eº_cell - RT/nF ln Q

So, we must first find Q:

Q = [Zn²⁺]/[Cu²⁺] = 1.0 / 1.0x10^-3 = 1x10³

Simplifying the Nernst equation assuming the temperature is 25ºC, we have...

E = Eº - 0.0592/n log₁₀ Q where n = mole electrons = 2

E = 1.10 - (0.0592/2) log 10³

E = 1.10 - 0.0888

E = 1.01 V