Determine E°cell for the reaction: Pb²⁺ + Zn → Pb + Zn²⁺. The half reactions are: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V
Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Question

Determine E°cell for the reaction: Pb²⁺ + Zn → Pb + Zn²⁺. The half reactions are: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Quizplus · Accepted Answer

To determine E°cell we need to use the formula: E°cell = E°cathode - E°anode Where E°cathode is the standard reduction potential of the cathode (reduction half reaction) and E°anode is the standard reduction potential of the anode (oxidation half reaction). In this case, the reduction half reaction is: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V And the oxidation half reaction is: Zn(s) → Zn²⁺(aq) + 2 e- E° = -0.763 V Notice that we need to reverse the oxidation half reaction to obtain a reduction half reaction: Zn²⁺(aq) + 2 e- → Zn(s) E° = +0.763 V Now we can calculate E°cell: E°cell = E°cathode - E°anode E°cell = -0.125 V - (+0.763 V) E°cell = -0.888 V However, this value is negative because the reaction is not spontaneous under standard conditions. Therefore, we need to reverse the sign of E°cell: E°cell = +0.888 V Therefore, the best choice is A, with an explanation that the value of E°cell is calculated by subtracting the E° of the anode from the E° of the cathode, and in this case, the answer is +0.888 V.

Determine E°cell for the Reaction: Pb2+ + Zn → Pb

Determine E°cell for the Reaction: Pb²⁺ + Zn → Pb