The quantity of electricity in Faraday needed to reduce $$1$$ mol of $$Cr_2O_7^{2-}$$ to $$Cr^{3+}$$ is ______

We need to find the quantity of electricity (in Faraday) required to reduce 1 mol of $$Cr_2O_7^{2-}$$ to $$Cr^{3+}$$.

First, determine the oxidation states: in $$Cr_2O_7^{2-}$$ each Cr is +6, while in $$Cr^{3+}$$ it is +3. Therefore each Cr atom gains 3 electrons according to

$$Cr^{+6} + 3e^- \to Cr^{+3}$$.

Since each $$Cr_2O_7^{2-}$$ ion contains 2 Cr atoms, the total number of electrons required per formula unit is $$2 \times 3 = 6\text{ electrons}$$. The balanced half-reaction is:

$$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$$.

Because one Faraday corresponds to one mole of electrons, reducing one mole of $$Cr_2O_7^{2-}$$ requires $$6\text{ Faraday}$$. Therefore, the answer is 6.