How much charge is required for the following reductions?

(i) $1 m o l$ of $A l^{3 +}$ to $A l$
(ii) $1 m o l$ of $C u^{2 +}$ to $C u$
(iii) $1 m o l$ of ${M n O_{4}}^{-}$ to $M n^{2 +}$

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Solution

$A l^{3 +} + 3 e^{-} \to A l$
The reduction will require 3 Faraday of electricity or $3 \times 96500 = 2.895 \times 10^{5} C$ .

$C u^{2 +} + 2 e^{-} \to C u$
The reduction will require 2 Faraday of electricity or $2 \times 96500 = 1.93 \times 10^{5} C$ .

${M n O_{4}}^{-} + 5 e^{-} \to M n^{2 +}$
The reduction will require 5 Faraday of electricity or $5 \times 96500 = 4.825 \times 10^{5} C$ .

Note: 1 mole of electron has a charge of 1 faraday.

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Similar questions

How much charge is required for the following reductions:

(i) 1 mol of Al³⁺ to Al.

(ii) 1 mol of Cu²⁺to Cu.

(iii) 1 mol of to Mn²⁺.

How much charge is required for the following reductions :

(i) 1 mole of $A l^{3 +}$ to Al ?

(ii) 1 mole of $C u^{2 +}$ to Cu ?

(iii) 1 mole of $M n O_{4}^{-} t o M n^{2 +}$ ?

1 m o l e

A l^{3 +}

A l

1 m o l e

M n O_{4}^{-}

M n^{2 +}

1 m o l

{M n O_{4}}^{-}

M n O_{2}

1 m o l e

A g^{+}

A g

1 m o l e

C u^{2 +}

C u

1 m o l e

M n O_{4}^{-}

M n O_{4}^{2 -}

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