Consider a galvanic cell based upon the following half reactions:


Screenshot 2024 01 24 000032

Screenshot 2024 01 24 000032

Step 1 of 2

Given data: Half reactions:

Pb+2e−⟶Pb−0.13V

Au3++3e−⟶Au 1.50V

Find: How would the following changes alter the potential of the cell?

 

  1. Adding Au ions to the gold half reaction (assume no volume change)
  2. Removing Pb2+ ions from solution by precipitating them out of the lead half reaction
  3. Adding equal amounts of water to both half reactions.
  4. Adding Pb ions to the lead half reaction (assuming no volume change)

Explanation:

The cell potential (E) for a galvanic cell can be calculated using the Nernst equation:

E=E°−(RT / (�F))⋅ln(Q)

 

Step 2 of 2

To determine how the given changes would alter the potential of the cell, we need to consider the effect of concentration changes on the cell potential.

The cell potential (E) for a galvanic cell can be calculated using the Nernst equation:

E=E°−(RT / (�F))⋅ln(Q)

Where:

E = cell potential

E° = standard cell potential

R = gas constant (8.314 J/(mol K)) T = temperature (in Kelvin)

n = number of electrons transferred in the balanced equation

F = Faraday’s constant (96485 C/mol)

ln = natural logarithm

Q = reaction quotient (concentration of products / concentration of reactants)

 

Now let’s analyze each change and its effect on the cell potential:

  1. Adding Au ions to the gold half-reaction (assume no volume change):
    • Adding Au ions to the gold half-reaction would shift the equilibrium towards the right, favoring the reduction of Au3+to Au. This results in an increase in the concentration of Au3+ and the reduction reaction. Consequently, the potential of the gold half-reaction would increase.

 

  1. Removing Pb2+ ions from solution by precipitating them out of the lead half-reaction (assume no volume change):
    • If Pb2+ ions are removed from the solution, the equilibrium in the lead half-reaction would shift to the left, favoring the oxidation of Pb to Pb2+. This results in a decrease in the concentration of Pb2+ and the oxidation reaction. Consequently, the potential of the lead half-reaction would decrease.

 

  1. Adding equal amounts of water to both half-reactions:
    • Adding water doesn’t affect the concentrations of ions involved in the half-reactions. The concentration of water is essentially constant, and it does not appear in the equilibrium expression. Therefore, the addition of water does not alter the potential of either half-reaction.

 

  1. Adding Pb ions to the lead half-reaction (assuming no volume change):
    • Adding Pb ions to the lead half-reaction would shift the equilibrium towards the right, favoring the reduction of Pb2+ to Pb. This results in an increase in the concentration of Pb2+ and the reduction reaction. Consequently, the potential of the lead half-reaction would increase.

Explanation:

Hence using Nernst equation we are able to determine the following above.

Final solution

In summary:

  • Adding Au ions to the gold half-reaction increases its potential.
  • Removing Pb2+ ions from the lead half-reaction decreases its potential.
  • Adding water to both half-reactions does not alter their potentials.
  • Adding Pb ions to the lead half-reaction increases its potential.

 

 

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