Question 17.7: EXPLORING HOW CHANGES IN CONCENTRATIONS AFFECT CELL VOLTAGE ...
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 tells us that lead is the anode and silver is the cathode. Therefore, the cell reaction is Pb(s) + 2 Ag^{+}(aq) → Pb^{2+}(aq) + 2 Ag(s).
The cell potential at 25 °C is given by the Nernst equation, where n = 2 and Q = [Pb^{2+}]/[Ag^{+}]^{2}
E = E° – \frac{0.0592 V}{n} log Q
= E° – \left(\frac{0.0592 V}{2}\right)\left(log \frac{[Pb^{2+}]}{[Ag^{+}]^{2}}\right)
The change in E on changing the ion concentrations will be determined by the change in the log term in the Nernst equation.
(a) Pb^{2+} is in the anode compartment, and Ag^{+} is in the cathode compartment. Suppose that the original concentrations of Pb^{2+} and Ag^{+} are 1 M, so that E = E°.
Increasing [Pb^{2+}] to 10 M gives
E = E° – \left(\frac{0.0592 V}{2}\right)\left(log\frac{(10)}{(1)^{2}}\right)
Because log 10 = 1.0, E = E° – 0.03 V. Thus, increasing the Pb^{2+} concentration by a factor of 10 decreases the cell voltage by 0.03 V.
(b) Increasing [Ag^{+}] to 10 M gives
E = E° – \left(\frac{0.0592 V}{2}\right)\left(log \frac{(1)}{(10)^{2}}\right)
Because log (10)^{-2} = -2.0, E = E° + 0.06 V. Thus, increasing the Ag^{+} concentration by a factor of 10 increases the cell voltage by 0.06 V.
BALLPARK CHECK
We expect that the reaction will have a lesser tendency to occur when the product ion concentration, [Pb^{2+}] , is increased and a greater tendency to occur when the reactant ion concentration, [Ag^{+}], is increased. Therefore, the cell voltage will decrease when [Pb^{2+}] is increased and will increase when [Ag^{+}] is increased. The ballpark check and the solution agree.