Calculate the electrode potential at 25°C of Cr³⁺, Cr₂O₇^2- electrode at pOH = 11 in a solution of 0.01 M both in Cr³⁺ and Cr₂O₇^2–. E° value for the cell

\(Cr_2O_7^{-2} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\)

E° = 1.33 V.

0.936 V

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, \(E_{cell}  = 0.936 V\)