Calculate the electrode potential at 25°C of Cr3+, Cr2O72- electrode at pOH = 11 in a solution of 0.01 M both in Cr3+ and Cr2O72–. E° value for the cell
\(Cr_2O_7^{-2} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\)
E° = 1.33 V.
Answer & explanation
Correct answer: option 2
The given cell reaction is
\(\underset{0.01}{Cr_2O_7^{2−}} + \underset{10^{−3}}{14H^+} + 6e^− \rightarrow \underset{0.01}{2Cr^{3+} + 7H_2O}\) \(E^0 = 1.33 V\)
Given,
\(pOH = 11\)
\(pH = 3\)
\([H^+] = 10^{−3}\)
\(E_{cell} = E^0_{cell} − \frac{0.0591}{6} log\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2−}][H^+]^{14}}\)
or, \(E_{cell} = 1.33 − \frac{0.0591}{6} log\frac{[0.01]^2}{[0.01][10^{−3}]}\)
or, \(E_{cell} = 0.936 V\)