Question 8

The reduction half-reaction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ From the stoichiometry, 6 moles of electrons are required to produce 2 moles of $Cr^{3+}$. To produce 1 mole of $Cr^{3+}$, the moles of electrons required ($n$) = 3. Total charge $Q = n \times F = 3 \times 96500\ C = 289500\ C$. Time $t = 48.25\ \text{minutes} = 48.25 \times 60\ \text{seconds} = 2895\ s$. Using $Q = I \times t$, we get $I = \frac{Q}{t} = \frac{289500}{2895} = 100\ A$.