Consider this cell: Cu(s) | Cu+(sq) || Mn+2(aq) | Mn (s) Write balanced half-rea
ID: 836250 • Letter: C
Question
Consider this cell: Cu(s) | Cu+(sq) || Mn+2(aq) | Mn (s) Write balanced half-reactions and the net reaction, (b) label each half-reaction as either "oxidation" or "reduction", and (c) calculate delta G degree for the reaction. Cu+ + e- equivalent Cu(s) Mn+2 + 2e- equivalent Mn (s) E degree = 0.518 V E degree = -1.182 V Calculate the voltage of the cell described in Question 6 operating under the following conditions: 0.020 M MnSo4 in the manganese half-cell and 0.50 M CuNo3 in the copper half-cell.Explanation / Answer
Question 1
First, write half-rxns:
Oxidation (occurs at anode and is to the left of the salt bridge): Cu(s) = Cu^+(aq) + 1e^-
Reduction (occurs at cathode and is to the right of the salt bridge): Mn^2+(aq) + 2e^- = Mn(s)
Balance the electrons in both half-rxns such that they cancel when both half-rxns are combined to give an overall cell reaction:
The first half-rxn gets multiplied by 2 and the second one stays as is.
2(Cu(s) = Cu^+(aq) + 1e^-)) becomes 2Cu(s) = 2Cu^+(aq) + 2e^-
n = 2 electrons have been tranferred from the anode to the cathode
Balanced half-rxns:
Oxidation: 2Cu(s) = 2Cu^+(aq) + 2e^-
Reduction: Mn^2+(aq) + 2e^- = Mn(s)
Combine both half-rxns to give the net reaction:
2Cu(s) + Mn^2+(aq) + 2e^- = 2Cu^+(aq) + Mn(s) + 2e^- ... electrons MUST cancel
2Cu(s) + Mn^2+(aq) = 2Cu^+(aq) + Mn(s) ... this is the net reaction
Solve for E^o(cell):
E^o(cell) = E^o(cathode) - E^o(anode) = -1.182 V - 0.518 V = -1.7 V = -1.7 J/C
Solve for delta G^o:
delta G^o = -nFE = -(2 moles of electrons)(96,485 C/mole)(-1.7 J/C) = -3.28 x 10^5 J = -328 kJ
Question 2
Use the Nernst equation to solve for E(cell):
E(cell) = E^o(cell) - (0.0592 / n) log Q
Q = [Cu^2+]^2 / [Mn^2+] = (0.50)^2 / 0.020 = 12.5
E(cell) = -1.7 - (0.0592 / 2) log 12.5 = -1.73 V