Current Required for Reduction of Dichromate to Chromium Ion

The balanced half-reaction for the reduction of dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) to chromium(III) ion ($\text{Cr}^{3+}$) in an acidic medium is:

$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \longrightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$

From the stoichiometry of the reaction:

• The production of $2\text{ moles}$ of $\text{Cr}^{3+}$ requires $6\text{ moles}$ of electrons ($6\text{ F}$ of charge).
• Therefore, the production of $1\text{ mole}$ of $\text{Cr}^{3+}$ requires: $n = \frac{6}{2} = 3\text{ moles of electrons} = 3\text{ F}$

The total charge $Q$ required is: $Q = 3 \times 96500\text{ C} = 289500\text{ C}$

The time $t$ for which the current flows is:

$$t = 48.25\text{ minutes} = 48.25 \times 60\text{ s} = 2895\text{ s}$$

Using the relation $Q = I \times t$, the current $I$ is calculated as:

$$I = \frac{Q}{t} = \frac{289500\text{ C}}{2895\text{ s}} = 100\text{ A}$$

Thus, the current that flows through the cell is $100\text{ A}$.