Consider the reaction : $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ .

What is the quantity of electricity in coulombs needed to reduce 1 mole of $C r_{2} O_{7}^{2 -}$ ?

579000 C

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679000 C

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879000 C

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979000 C

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Solution

The correct option is A $579000 C$
${C r}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {C r}^{3 +} + 7 H_{2} O$

For complete reduction of $1$ mole of ${C r}_{2} O_{7}^{2 -}, 6 e^{-}$ is needed.

Hence, Quantity of electricity $= 6 \times 96500 = 579000$ $C$ .

Hence, option A is correct.

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C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-}

\to

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C r_{2} O_{7}^{2 -}

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C r_{2} O_{7}^{2 -}

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