1. A voltaic cell is constructed with a Cd metal strip in 0.07 M Cd(2+) in one h
ID: 537088 • Letter: 1
Question
1. A voltaic cell is constructed with a Cd metal strip in 0.07 M Cd(2+) in one half cell, and a Cu metal strip in 0.70 M Cu(2+) in the other half cell, in the same configuration as we saw in the lecture notes.
a. Search for the standard reduction potentials of these two half reactions and state them with the written half reaction.
b. Write the balanced equation for the voltaic cell
c. Determine the cell potential under standard conditions (25 degrees celsius, and all concentrations at 1M).
d. Determine the cell potential (emf) with the concentrations in the voltaic cell listed above.
e. Could Cd be used for the cathodic protection of Cu ? Explain your answer by reference to the standard reduction potentials.
Explanation / Answer
(a)
E0Cd2+/Cd = - 0.403 V
E0Cu2+/Cu = + 0.337 V
Since, E0Cu2+/Cu > E0Cd2+/Cd,
Cd under goes oxidation and Cu2+ undergoes reduction.
Oxidation Half Reaction: Cd (s) --------------> Cd2+ (aq.) + 2 e
Reduction Half Reaction: Cu2+ (aq.) + 2 e ---------------> Cu (s)
(b) Balanced cell reaction is,
Cd (s) + Cu2+ (aq.) ---------> Cd2+ (aq.) + Cu (s)
(c) Standar cell potential,
E0cell = E0Cu2+/Cu - E0Cd2+/Cd = + 0.337 - ( - 0.403 ) = + 0.740 V
(d)
Q = [Cd2+] / [Cu2+] = 0.07 / 0.70 = 0.1
Applying Nernst equation for cell reaction,
Ecell = E0cell - (0.05916 / n) LogQ
Ecell = 0.740 - (0.05916 / 2 ) Log(0.1)
Ecell = 0.770 V
(e)
Yes. Because the standard reduction potential of Cd2+ is smaller than that of Cu2+. Hence Cd is ready to lose electrons which can be helpful to protect the Cu from oxidation.
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