Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.)Pb(s) + Br2(l) → Pb2+(aq) + 2 Br-(aq)Given the standard reduction potentials:Pb2+(aq) + 2 e- → Pb(s) E° = -0.13 VBr2(l) + 2 e- → 2 Br-(aq) E° = +1.07 VWhat is the standard cell potential (E°cell) for the reaction?

• A. 1.20 V
• B. 0.80 V
• C. 0.94 V
• D. 1.00 V

Answer: (A)

Verified step by step guidance

1. Identify the half-reactions involved in the overall reaction. The given half-reactions are: Pb2+(aq) + 2 e- → Pb(s) with E° = -0.13 V and Br2(l) + 2 e- → 2 Br-(aq) with E° = +1.07 V.
2. Determine which half-reaction is the oxidation and which is the reduction. In the overall reaction, Pb(s) is oxidized to Pb2+(aq), and Br2(l) is reduced to 2 Br-(aq).
3. Write the oxidation half-reaction: Pb(s) → Pb2+(aq) + 2 e-. The standard potential for this oxidation reaction is the negative of the reduction potential: E° = +0.13 V.
4. Write the reduction half-reaction: Br2(l) + 2 e- → 2 Br-(aq). The standard potential for this reduction reaction is given as E° = +1.07 V.
5. Calculate the standard cell potential (E°cell) by adding the standard potentials of the oxidation and reduction half-reactions: E°cell = E°(oxidation) + E°(reduction) = +0.13 V + 1.07 V.