Question

A voltaic cell consists of a copper electrode in a solution of copper(II) ionsand a palladium electrode in a solution of palladium(II) ions. The palladiumis the cathode and its reduction potential is 0.951 V.

(a) Write the half-reaction that occurs at the anode.
(b) If E° is 0.609 V, what is the potentialfor the oxidation half-reaction?
(c) What is Keqfor this reaction?

Answer 1

a)     Cu(s) -> Cu²⁺(aq) + 2e^-

b)     E°_cell = E°_red + E°_ox

0.609 V = 0.951 V + E°_ox

E°_ox = -0.340 V

-0.340 V

c)     log K_eq = n * E°/0.0592

log K_eq = 2 * 0.609 V/0.0592 = 20.6

K_eq = 10^20.6 = 3.98 * 10²⁰

3.98 * 10²⁰