Question

(6): Calculate the maximum amount of work that can be obtained from the galvanic cells at standard conditions listed below. (Use the table for standard reduction potentials.)

(a) Cr2O72− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2O

H2O2 + 2 H+ + 2 e− → 2 H2O


____ kJ



(b) 2 H+ + 2 e− → H2

Al3+ + 3e− → Al


____ kJ

Answer 1

(a) The maximum amount of work that can be obtained from the first galvanic cell is -510.5kJ.

(b) The maximum amount of work that can be obtained from the second galvanic cell is -318.5kJ.

Step-by-step explanation:

In this question, we need to use the standard reduction potentials to calculate the standard cell potential of each galvanic cell, and then use the formula AG = -nFE to calculate the maximum amount of work that can be obtained from each galvanic cell.

Solve the Problem

(a) Cr₂02 +14H++6e → 2Cr³++7H₂O

The standard reduction potential of Cr2O2 is 1.33V, and the standard reduction potential of His 0V. Therefore, the standard cell potential of this galvanic cell is:

E cell = reduction (Cr202) - E reduction (H+) = 1.33V

The number of electrons transferred in this reaction is 6, so the maximum amount of work that can be obtained from this galvanic cell is:

AG= -nFE = -696485C/mol 1.33J/C = −51

(b) 2H+ + 2e→ H₂

The standard reduction potential of H+ is 0V, and the standard reduction potential of Al3+ is -1.66V.

Therefore, the standard cell potential of this galvanic cell is:

EO =E0 cell reduction (H+)-E0 reduction (A13+) = 0V – (−1

The number of electrons transferred in this reaction is 2, so the maximum amount of work that can be obtained from this galvanic cell is:

AG-nFE-2 96485C/mol 1.66J/C-31

Draw the Conclusion

(a) The maximum amount of work that can be obtained from the first galvanic cell is -510.5kJ.

(b) The maximum amount of work that can be obtained from the second galvanic cell is -318.5kJ.