Calculate the maximum amount of work that can be obtained from the galvanic cells at standard conditions in Exercise 45.

Interpretation:

The set up of two galvanic cells is given. The value of maximum work that can be obtained from the given galvanic cell is to be calculated.

Concept introduction:

The species with higher reduction potential undergoes reduction at the cathode while the species with lower reduction potential undergoes oxidation at the anode. The change in Gibbs free energy is the useful amount work that can be done and negative value of ΔG° shows that the reaction is spontaneous in nature.

To determine: The value of maximum work that can be obtained from the given galvanic cell.

The value of maximum work that can be obtained from the given galvanic cell is calculated as -387.87 kJ_ .

The reaction taking place at cathode is,

Au3+(aq)+3e−→Au(s)   E°red=+1.50 V

The reaction taking place at anode is,

Cu+(aq)→Cu2+(aq)+e−   E°ox=−0.16 V

Multiply the oxidation half reaction with a coefficient of 3 and then add both the oxidation and reduction half-reaction,

Au3+(aq)+3e−→Au(s)3Cu+(aq)→3Cu2+(aq)+3e−

The final equation is,

Au3+(aq)+3Cu+(aq)→Au(s)+3Cu2+(aq)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=−0.16 V+1.50 V=1

Interpretation:

The set up of two galvanic cells is given. The value of maximum work that can be obtained from the given galvanic cell is to be calculated.

Concept introduction:

The species with higher reduction potential undergoes reduction at the cathode while the species with lower reduction potential undergoes oxidation at the anode. The change in Gibbs free energy is the useful amount work that can be done and negative value of ΔG° shows that the reaction is spontaneous in nature.

To determine: The value of maximum work that can be obtained from the given galvanic cell.