In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to Cr$$^{3+}$$. The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of Cr$$^{3+}$$ is ________.

Use: 1 Faraday = 96500 C mol$$^{-1}$$

The half-reaction for the reduction of dichromate ions in acidic medium is: $$Cr_2O_7^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_2O$$.

From the stoichiometry, $$6$$ electrons (or $$6$$ Faradays) produce $$2$$ moles of $$Cr^{3+}$$.
Therefore, electrons required per mole of $$Cr^{3+}$$:

$$\frac{6\text{ mol e}^{-}}{2\text{ mol }Cr^{3+}}=3\text{ mol e}^{-}\text{ per mol }Cr^{3+}$$.

To obtain $$1$$ mole of $$Cr^{3+}$$, charge needed is:
$$Q = 3F = 3\times 96500\ \text{C mol}^{-1}=289500\ \text{C}$$.

The current flows for $$48.25\ \text{min}$$:
$$t = 48.25\times 60 = 2895\ \text{s}$$.

Current is $$I = \dfrac{Q}{t} = \dfrac{289500\ \text{C}}{2895\ \text{s}} = 100\ \text{A}$$.

Hence, the required current is 100 A.