Exhibit 18-3 Use the standard reduction potentials below to answer the following question(s) .
Ce⁴⁺(aq) + e^-→Ce³⁺(aq) E ° = 1.61 V
Fe³⁺(aq) + e^-→Fe²⁺(aq) E ° = 0.77 V
I₂(s) + 2 e^-→2 I^-(aq) E ° = 0.54 V
2 H⁺(aq) + 2 e^-→H₂(g) E ° = 0.000 V
Cd²⁺(aq) + 2 e^-→Cd(s) E ° = - 0.40 V
U⁴⁺(aq) + e^-→U³⁺(aq) E ° = - 0.61 V
CdCO₃(s) + 2 e^-→Cd(s) + CO₃^{2 -}(aq) E ° = - 0.74 V
Zn²⁺(aq) + 2 e^-→Zn(s) E ° = - 0.76 V
U³⁺(aq) + 3 e^-→U(s) E ° = - 1.80 V
Mg²⁺(aq) + 2 e^-→Mg(s) E ° = - 2.37 V
-Refer to Exhibit 18-3. Find the standard potential for the reaction:
2 U⁴⁺ + 2 I^-→2 U³⁺ + I₂ (s)

Question

Quizplus · Accepted Answer

First, we need to break down the given reaction into the individual half-reactions and find their standard potentials:
2 USO4 2- + 3e- → 2 U  (E° = -1.80 V)
2 I- → I2 + 2e- (E° = +0.54 V)
Hence, the overall reaction can be written as:
2 USO4 2- + I2 + 2e- → 2 U  + 2 USO4 2- + I- (E° = -1.26 V)
Now, we need to add another half-reaction to form solid ISb:
CdSbO3 + 4e- → Cd + Sb (E° = -0.74 V)

Combining the two half reactions:
2 USO4 2- + I2 + CdSbO3 + 4e- → 2 U  + 2 USO4 2- + I- + Cd + Sb (E° = -2.00 V)

Now, we need to reverse the second half-reaction to create solid ISb, and reverse the entire reaction:
2 U  + 2 USO4 2- + I- + Cd + Sb → 2 USO4 2- + I2 + CdSbO3 + 4e- (E° = +2.00 V)

Therefore, the standard potential for the given reaction is +2.00 V. 
Since the answer choices are not given in volts, we must convert to the proper format:
+2.00 V = 2.00 volts
Therefore, the closest answer is D (-1.15 V), but none of the given answer choices are exactly correct.

Exhibit 18-3 Use the Standard Reduction Potentials Below to Answer