Calculate for an electrochemical cell reaction that occurs under acidic aqueous conditions based on the following two half-reactions for which the standard reduction potentials are given.Use the smallest whole-number coefficients possible when balancing the overall reaction. Cr₂O₇^2- + 14 H⁺ + 6 e^-$\to$ 2 Cr³⁺ + 7 H₂O $\quad $ $\quad $+1.33 V CH₃COOH + 4 H⁺ + 4 e^-$\to$ C₂H₅OH + H₂O $\quad $+0.058 V

Question

Quizplus · Accepted Answer

1. Write the half-reactions and their standard reduction potentials:
Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O , E° = 1.33 V 
4H+ + 4e− + CH3COOH → CH4 + HCOO− , E° = 0.058 V 
2. Balance the electrons and combine the half-reactions:
2Cr2O72− + 14H+ + 12e− → 4Cr3+ + 14H2O 
3CH3COOH + 12H+ + 12e− → 3CH4 + 3HCOO− 
Multiplying the second half reaction by 2, we get:
6CH3COOH + 24H+ + 24e− → 6CH4 + 6HCOO− 
4. Combine the half reactions to obtain the balanced overall cell reaction:
2Cr2O72− + 6CH3COOH + 28H+ → 4Cr3+ + 6CH4 + 14H2O + 6HCOO− 
5. Calculate the standard cell potential:
E°cell = E°reduction (cathode) - E°reduction (anode) = 0.058 V - 1.33 V = -1.272 V 
6. The standard Gibbs free energy change (ΔG°) is related to the cell potential (E°cell) by the equation:
ΔG° = -nFE°cell 
where n is the number of electrons transferred in the cell reaction, F is the Faraday constant (96,485 C/mol), and the negative sign indicates a non-spontaneous reaction. For the given reaction, n = 12, so:
ΔG° = -12 x 96,485 C/mol x (-1.272 V) = +1.47 x 10^5 J/mol 
Converting to kJ/mol and using proper scientific notation, we get:
ΔG° = +147 kJ/mol ≈ +1470 kJ/mol ≈ +1470 kJ 
Therefore, the best choice is C, which is the closest approximation to +1470 kJ.

Calculate for an Electrochemical Cell Reaction That Occurs Under →\to→

Calculate for an Electrochemical Cell Reaction That Occurs Under $\to$