The charge required for reduction of $1 m o l$ of $C r_{2} O_{7}^{2 -}$ ions to $C r^{3 +}$ is:

9650 C

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2 \times 96500 C

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3 \times 96500 C

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6 \times 96500 C

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Solution

The correct option is A $6 \times 96500 C$
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 8 H_{2} O$

From the balanced reaction, we get $n = 6$

Charge required $= n F = 6 \times 96500 C = 579000 C$

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In electrolytic reduction of a nitroarene with 50% current efficiency, 20.50 g of the compound is reduced by $2 \times 96500$ C of electric charge. The molar mass of the compound is :

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The charge required for reduction of 1 mol of Cr2O2−7 ions to Cr3+ is:

The correct option is A 6×96500CCr2O2−7+14H++6e−→2Cr3++8H2OFrom the balanced reaction, we get n=6Charge required =nF=6×96500C=579000C

The charge required for reduction of $1 m o l$ of $C r_{2} O_{7}^{2 -}$ ions to $C r^{3 +}$ is:

The correct option is A $6 \times 96500 C$
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 8 H_{2} O$

From the balanced reaction, we get $n = 6$

Charge required $= n F = 6 \times 96500 C = 579000 C$