Exploring How Changes in Concentrations Affect Cell Voltage Consider the following galvanic cell: (a) What is the change in the cell voltage on increasing the ion concentrations in the anode compartment by a factor of 10?(b) What is the change in the cell voltage on increasing the ion

Question

Exploring How Changes in Concentrations Affect Cell Voltage
Consider the following galvanic cell:

(a) What is the change in the cell voltage on increasing the ion concentrations in the anode compartment by a factor of 10?
(b) What is the change in the cell voltage on increasing the ion concentrations in the cathode compartment by a factor of 10?

STRATEGY
The direction of electron flow in the picture indicates that lead is the anode and silver is the cathode. Therefore, the cell reaction is

[latex]Pb(s) + 2 Ag^{+}(aq) → Pb^{2+}(aq) + 2 Ag(s)[/latex]

The cell potential at 25 °C is given by the Nernst equation, where n = 2 and [latex]Q = [Pb^{2+}]/[Ag^{+}]^{2}[/latex] :

[latex]E = E° -\frac{0.0592 V}{n} \log Q = E° - (\frac{0.0592 V}{2}) (\log\frac{[Pb^{2+}]}{[Ag^{+}]^{2}})[/latex]

The change in E on changing the ion concentrations will be determined by the change in the log term in the Nernst equation.

Accepted Answer

(a) [latex]Pb^{2+}[/latex] is in the anode compartment, and [latex]Ag^{+}[/latex] is in the cathode compartment. Suppose that the original concentrations of [latex]Pb^{2+}[/latex] and [latex]Ag^{+}[/latex] are 1 M, so that E = E°. Increasing [latex][Pb^{2+}][/latex] to 10 M gives

[latex]E = E° -(\frac{0.0592 V}{2})\log \frac{(10) }{(1)^{2}})[/latex]

Because log 10 = 1.0, E = E° - 0.03 V. Thus, increasing the [latex]Pb^{2+}[/latex] concentration by a factor of 10 decreases the cell voltage by 0.03 V

(b) Increasing [latex][Ag^{+}][/latex] to 10 M gives

[latex]E = E° -(\frac{0.0592 V}{2})\log\frac{(1)}{(10)^{2}})[/latex]

Because [latex]\log (10)^{-2} = -2.0, E = E° + 0.06 V.[/latex] Thus, increasing the [latex]Ag^{+}[/latex] concentration by a factor of 10 increases the cell voltage by 0.06 V.

CHECK

We expect that the reaction will have a lesser tendency to occur when the product ion concentration, [latex][Pb^{2+}][/latex], is increased and a greater tendency to occur when the reactant ion concentration, [latex][Ag^{+}][/latex], is increased. Therefore, the cell voltage E will decrease when [latex][Pb^{2+}][/latex] is increased and will increase when [latex][Ag^{+}][/latex] is increased. The prediction and the solution agree.

Known	Unknown
Concentration changes	Change in E
Cell diagram (used to determine overall reaction)

Question 19.10: Exploring How Changes in Concentrations Affect Cell Voltage ...

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