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Faraday's First Law

How much char...

Question

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 $M n O_{4}^{-} t o M n^{2 +}$ ?

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Solution

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

Electrode reaction : $A l^{3 +} 1 m o l (a q) + 3 e^{-} 3 m o l \to A l (s)$
Quantity of charge required for reduction of 1 mole of $A l^{3 +} = 3 F a r a d a y = 3 \times 96500 C = 2.895 \times 10^{5} C$

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

Electrode reaction : $C u^{2 +} 1 m o l (a q) + 2 e^{-} 2 m o l \to C u (s)$
Quantity of charge required for reducing 1 mole $C u^{2 +} t o C u = 2 F = 2 \times 96500 C = 1.93 \times 10^{5} C$

(iii)

$1 m o l e M n O_{4}^{-} t o M n^{2 +}$

$E l e c t r o d e r e a c t i o n : M n O_{4}^{-} \begin{matrix} o x . n o o f M n = 7 1 m o l \end{matrix} + 5 e^{-} 5 m o l \to M n O_{4}^{-} \begin{matrix} (o x . n o o f M n = 2) \end{matrix}$

Quantity of charge required for reducing 1 mole $M n O_{4}^{-}$ to $M n^{2 +} = 5 F = 5 \times 96500 = 4.825 \times 10^{5} C$

Suggest Corrections

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⁺.

1 m o l

A l^{3 +}

A l

1 m o l

C u^{2 +}

C u

1 m o l

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

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 +}

V O_{3}^{2 -} + M n O_{4}^{-} ⟶ {M n}^{2 +} + V O_{4}^{3 -}

M n O_{4}

1

V O_{3}^{2 -}

A g^{\oplus}

\frac{1}{2}

C u^{2 +}

\frac{1}{3}

A l^{3 +}

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