The quantity of electricity in Faraday needed to reduce 1 mol of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ to $\mathrm{Cr}^{3+}$ is ______.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Redox Reactions and Electrochemistry

The quantity of electricity in Faraday needed to reduce 1 mol of

Cr_{2} O_{7}^{2 -}

Cr^{3 +}

is ______.

Answer: 6

Solution

The reduction half-reaction is:

Cr_{2} O_{7}^{2 -} + 6 e^{-} \to 2 Cr^{3 +}

Each chromium atom goes from oxidation state +6 (in dichromate) to +3, requiring 3 electrons per Cr atom. For 2 Cr atoms, 6 electrons are needed.

Therefore, 1 mole of

Cr_{2} O_{7}^{2 -}

requires 6 moles of electrons, which equals 6 Faradays of electricity.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer

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