Electrochemistry

Question

Given that Zn/Zn2+(1M)//Cu2+(1M)/Cu

(a) calculate the emf of the cell

(b) predict if the reaction is spontaneous or non spontaneous

Given that Zn/Zn2+ E=0.34V, Cu2+/Cu E=1.10V

Answer 1

Zn²⁺ + 2e- ==> Zn Eº = 0.34 V (anode)

Cu²⁺ + 2e- ==> Cu Eº = 1.10 V (cathode)

Overall reaction:

Cu²⁺ + Zn ==> Cu + Zn²⁺

Eº = cathode - anode = 1.10 - 0.34 = 0.76 V = emf of the cell

Is this spontaneous? YES, because the emf is POSITIVE.