Question

Given \[\ce{E^{\circ}_{Cr^{3+}/Cr}}\] = −0.72 V, \[\ce{E^{\circ}_{Fe^{2+}/Fe}}\] = −0.42 V. The potential for the cell \[\ce{Cr | Cr^{3+} (0.1 M) || Fe^{2+} (0.01 M) | Fe}\] is:

Options

• 0.26 V

• 0.339 V

• −0.339 V

• −0.26 V

Answer 1

0.26 V

Explanation:

\[\ce{E^{\circ}_{cell} = E^{\circ}_{R} - E^{\circ}_{L}}\]

= \[\ce{E^{\circ}_{Fe^{2+}/Fe} - E^{\circ}_{Cr^{3+}/Cr}}\]

= −0.42 − (−0.72)

= 0.3 V

The cell reaction for the given cell is:

\[\ce{2Cr_{(s)} + 3Fe^{2+}_{ (aq)} -> 2Cr^{3+}_{ (aq)} + 3Fe_{(s)}}\]

and n = 6

\[\ce{E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} log_10 \frac{[Cr^{2+}]^2}{[Fe^{2+}]^3}}\]

= \[\ce{0.3 - \frac{0.0591}{6} log_10 \frac{(0.1)^2}{(0.01)^3}}\]

= 0.3 − 0.039

= 0.261 V