The charge required for the reduction of 1 mole of Cr 2 O 72- to Cr 3- isA- 96500 C- 1-93 - 105 C- 5-79 - 105 CD- 2-895 - 105 C

The correct option is C 5.79 × 10^5CCr_2O_7^2 +6e^ → 2Cr^3+ Reduction of 1 mole of Cr_2O_7^2 TO Cr^3+ requires 6 mole of electrons. ∴ Charge required = 6 × ...

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Standard XII

Chemistry

Introduction to Atomic Structure

The charge re...

Question

The charge required for the reduction of 1 mole of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ is

96500 C

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1.93 \times 10^{5} C

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5.79 \times 10^{5} C

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2.895 \times 10^{5} C

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Solution

The correct option is C $5.79 \times 10^{5} C$
$C r_{2} O_{7}^{2 -} + 6 e^{-} \to 2 C r^{3 +}$
Reduction of 1 mole of $C r_{2} O_{7}^{2 -}$ TO $C r^{3 +}$ requires 6 mole of electrons.
$∴$ Charge required = 6 $\times$ 96500C
= 5.79 $\times 10^{5} C$

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The charge required for the reduction of 1 mole of Cr2O2−7 to Cr3+ is

The correct option is C 5.79×105CCr2O2−7+6e−→2Cr3+ Reduction of 1 mole of Cr2O2−7 TO Cr3+ requires 6 mole of electrons. ∴ Charge required = 6 × 96500C = 5.79 ×105C

The charge required for the reduction of 1 mole of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ is

The correct option is C $5.79 \times 10^{5} C$
$C r_{2} O_{7}^{2 -} + 6 e^{-} \to 2 C r^{3 +}$
Reduction of 1 mole of $C r_{2} O_{7}^{2 -}$ TO $C r^{3 +}$ requires 6 mole of electrons.
$∴$ Charge required = 6 $\times$ 96500C
= 5.79 $\times 10^{5} C$