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For each electrochemical cell, calculate the cell voltage using the table of standard reduction potentials linked in the instructions and indicate whether or not the reaction is spontaneous.(a) Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) VIs this reaction spontaneous as written?yesno    (b) 2Cu(s) + Br2(l) → 2Br-(aq) + 2Cu+(aq) VIs this reaction spontaneous as written?yesno    (c) 2Au(s) + 3MnO2(s) + 12H+(aq) → 2Au3+(aq) + 6H2O(l) + 3Mn2+(aq) V

Question

For each electrochemical cell, calculate the cell voltage using the table of standard reduction potentials linked in the instructions and indicate whether or not the reaction is spontaneous.(a) Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) VIs this reaction spontaneous as written?yesno    (b) 2Cu(s) + Br2(l) → 2Br-(aq) + 2Cu+(aq) VIs this reaction spontaneous as written?yesno    (c) 2Au(s) + 3MnO2(s) + 12H+(aq) → 2Au3+(aq) + 6H2O(l) + 3Mn2+(aq) V

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Solution

To answer this question, we need to use the Nernst equation, which is Ecell = E°cell - (RT/nF)lnQ. However, since we are dealing with standard conditions and Q=1, the equation simplifies to Ecell = E°cell.

(a) For the reaction Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq), we need to find the standard reduction potentials for the half-reactions.

Zn2+(aq) + 2e- → Zn(s) E° = -0.76 V Pb2+(aq) + 2e- → Pb(s) E° = -0.13 V

The cell potential is the difference between the reduction potential of the cathode and the anode. In this case, Pb is being oxidized and Zn is being reduced, so E°cell = E°cathode - E°anode = (-0.76 V) - (-0.13 V) = -0.63 V.

Since the cell potential is negative, the reaction is not spontaneous as written.

(b) For the reaction 2Cu(s) + Br2(l) → 2Br-(aq) + 2Cu+(aq), we find the standard reduction potentials for the half-reactions.

Cu+(aq) + e- → Cu(s) E° = 0.52 V Br2(l) + 2e- → 2Br-(aq) E° = 1.07 V

The cell potential is E°cell = E°cathode - E°anode = (1.07 V) - (0.52 V) = 0.55 V.

Since the cell potential is positive, the reaction is spontaneous as written.

(c) For the reaction 2Au(s) + 3MnO2(s) + 12H+(aq) → 2Au3+(aq) + 6H2O(l) + 3Mn2+(aq), we find the standard reduction potentials for the half-reactions.

Au3+(aq) + 3e- → Au(s) E° = 1.50 V MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l) E° = 1.23 V

The cell potential is E°cell = E°cathode - E°anode = (1.50 V) - (1.23 V) = 0.27 V.

Since the cell potential is positive, the reaction is spontaneous as written.

This problem has been solved

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