The Nernst equation: Consider the following galvanic cell
`Pb(s) | Pb^(2+)(aq.) || Ag^(2+)(aq.) | Ag(s)`
(a) What is the quantitative change in the cell voltage on increas-ing the ion concentrations in the anode compartment by a factor of `10`?
(b) What is the quantitative change in the cell voltage on increas-ing the ion concentration in the cathode compartment by a factor of `10`
Strategy: The conventional notation of the cell tells us that lead is the anode while silve is the cathode. Therfore, the cell reaction is
`Pb(s) + 2Ag^(+)(aq.) rarr Pb^(2+)(aq.) + 2Ag(s)`
The cell potantial (at `25^(@)C`) is given by the Nernst equation, where `n = 2` and `Q = C_(Pb^(2+))//C_(Ag^(+))`:
`E_("cell") = E_("cell")^(@)-(0.0592 V)/(n) "logQ"`
`E_("cell")^(@)-(0.0592 V)/(n) "logQ" C_(Pb^(2+))/C_(Ag^(+))^(2)`

Explanation

(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_("cell") = E_("cell")^(@)`
Increasing `C_(Pb^(2+))` to `10 M` gives
`E_("cell") = E_("cell")^(@) - ((0.0592 V)/(2))"log" (10)/(1^(2)`
or `E_("cell") = E_("cell")^(@) - 0.03 V` (because `"log" 10 = 1`)
Thus, increasing the `Pb^(2+)` concentration by a factor of `10` decreases the cell voltage by `0.03V`. (b) Increases `C_(Ag^(+))` to `10 M` gives
`E_("cell") = E_("cell")^(@) - ((0.0592 V)/(2))"log"(1)/((10)^(2))`
or `E_("cell") = E_("cell")^(@) + 0.06 V` (because `"log" 10^(-2) = -2.0`)
Thus, increasing the `Ag^(+)` concentration by a factor of `10` increases the cell voltage by `0.06V`.
The Nernst equation can also be used to convert measure potentials to standard potentials as illustrated by the following example.