For the cell reaction
Cr2O72− + 14H+ + 6I− → 2Cr3+ + 3I2 + 7H2O,
E°cell = 0.79 V and E°(Cr2O72−/Cr3+) = 1.33 V. What is E°(I2/I−)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → +0.54 V
In the cell reaction dichromate is reduced (cathode) and iodide is oxidised (anode):
Cathode: \( \displaystyle \ce{Cr2O7^2- + 14H^+ + 6e^- -> 2Cr^3+ + 7H2O} \)
Anode: \( \displaystyle \ce{6I^- -> 3I2 + 6e^-} \)
The cell potential is the electrode potential of the cathode minus that of the anode:
\( \displaystyle E^{\circ}_{\text{cell}}=E^{\circ}(\ce{Cr2O7^2-/Cr^3+})-E^{\circ}(\ce{I2/I^-}) \)
\( \displaystyle 0.79=1.33-E^{\circ}(\ce{I2/I^-}) \)
\( \displaystyle E^{\circ}(\ce{I2/I^-})=1.33-0.79=+0.54\,\mathrm{V} \)