Question

Calculate the e.m.f. of the following cell reaction at 298 K:

\[\ce{Mg_{(s)} + Cu{^{2+}} (0.0001 M) -> Mg{^{2+}} (0.001M) + Cu_{(s)}}\]

The standard potential (E°) of the cell is 2.71 V.

Answer 1

Given:

Cell reaction: \[\ce{Mg_{(s)} + Cu{^{2+}} (0.0001 M) -> Mg{^{2+}} (0.001M) + Cu_{(s)}}\]

Standard EMF, \[\ce{E{^{\circ}_{cell}}}\] = 2.71 V

Temperature (T) = 298 K

\[\ce{Q = \frac{Mg^{2+}}{Cu^{2+}}}\]

= \[\ce{\frac{10^{-3}}{10^{-4}}}\]

= 10

n = 2

\[\ce{E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} log Q}\]    ...(Using the Nernst equation)

= \[\ce{2.71 - \frac{0.0591}{2} log 10}\]

= 2.71 − 0.02955 × 1  ...(log 10 = 1)

= 2.71 − 0.02955

= 2.680 V