Question

A) Sketch the electrochemical cell made from Cu/Cu²⁺ and Ag/Ag⁺ half cells under standard conditions. (E⁰red (Ag⁺) = 0.80 V and E⁰red (Cu²⁺) = 0.34V)

B) Towards which half-cell do electrons flow in the external circuit?

C) In which half-cell does reduction occur?

D) Calculate E⁰cell in volts (show work).

E) Which electrode is the positive pole?

F) Which electrode decreases in mass during cell operation.

G) Which electrodes are considered active?

H) What differences, if any, in responses B, C, D, and E above would occur if the copper half cell was replaced with the SHE?

I) Write the overall balanced cell reaction for the copper/ silver redox.

J) Calculate ΔG⁰ for the reaction.

Answer 1

A) The half cell reactions are:

Ag⁺ (aq) + e^- -------> Ag (s); E⁰_red = 0.80 V

Cu²⁺ (aq) + 2 e^- ------> Cu (s); E⁰red = 0.34 V

The higher the electrode potential, the more oxidizing is the metal. Consequently, Ag⁺ can oxidize Cu to Cu²⁺. Consequently, Ag⁺ is reduced and Cu is oxidized. The electrochemical cell can be designated as below.

Cu (s) │Cu²⁺ (aq) ││Ag⁺ (aq)│Ag

B) The direction of flow of electrons in the external circuit is from the anode (oxidation half) to the cathode (reduction half).

C) Reduction takes place in the Ag⁺/Ag half cell and the reduction half reaction is

Reduction Half: Ag⁺ (aq) + e^- --------> Ag (s); E⁰_red = +0.80 V

D) The oxidation half reaction is

Oxidation Half: Cu (s) ------> Cu²⁺ (aq); E⁰_ox = -0.34 V

The cell voltage is given as

E⁰_cell = E⁰_red + E⁰_ox = (+0.80 V) + (-0.34 V) = +0.46 V (ans).

E) Cu/Cu²⁺ half cell comprises the oxidation half of the electrochemical cell.

F) The Cu metal is oxidized to Cu²⁺ while Ag⁺ from the electrolyte gets reduced to Ag. Consequently, the Cu electrode will reduce in mass.

G) The Cu electrode is itself oxidized to Cu²⁺ while Ag⁺ reduces to give Ag. Consequently, the Cu electrode is the active electrode.