A voltaic cell consists of a copper electrode in a solution of copper(II) ions w

Question

A voltaic cell consists of a copper electrode in a solution of copper(II) ions with unknown concentration,
and a palladium electrode in a 0.1 M solution of palladium(II) ions. The palladium electrode is the cathode
and its reduction potential is 0.951 V, the Cu electrode is the anode.

(a) Write the half-reaction that occurs at the anode and the cathode, as well as the overall cell reaction.

(b) What is the concentration of Cu2+ ions in the anode if the cell potential is 0.584 V? Speculate why this is
not a very accurate measurement to determine the Cu2+ concentration.

(c) What is the equilibrium constant (Keq) for this reaction?

Explanation / Answer

A) at anode Cu(s) --> Cu2+(aq) + 2e-
at cathod Pd2+ + 2e- --> Pd E°red = 0.951 V
________________________________________________________
overall cell Pd2+(aq) + Cu(s) ---> Pd(s) + Cu2+(aq) E°cell = 0.584 V
reaction

B) E°cell = E°red + E°ox
0.584 V = 0.951 V + E°ox
E°ox = -0.367 V

C) log Keq = n x E°/0.0592
log Keq = 2 x 0.584 V/0.0592 = 19.729
Keq = 10^19.729 = 5.357 x 10^19

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