Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method of preparation of MnO2 as per reaction- Mn2-aq-2H2O- MnO2s -2H-aq-H2gPassing a current of 27 A for 24 hr gives 1 kg of MnO2- What is the value of current efficiency-

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Byju's Answer

Standard XII

Chemistry

Faraday's Second Law

Electrolysis ...

Question

Electrolysis of a solution of $M n S O_{4}$ in aqueous sulphuric acid is a method of preparation of $M n O_{2}$ as per reaction, $M n^{2 +} (a q .) + 2 H_{2} O \to M n O_{2} (s) + 2 H^{+} (a q .) + H_{2} (g)$
Passing a current of 27 A for 24 hr gives 1 kg of $M n O_{2}$ . What is the value of current efficiency?

93.2%

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94.8%

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92.6%

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95.1%

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Solution

The correct option is D 95.1%
$w = \frac{E \times i \times t}{96500}$
$E (e q u i v a l e n t w e i g h t) = \frac{M o l e c u l a r w e i g h t}{c h a n g e i n o x i d a t i o n n o .}$
$E (M n O_{2}) = \frac{87}{2} = 43.5$
Thus, substituting values
$1000 = \frac{43.5 \times i \times 24 \times 3600}{96500}$
$i = 25.67 A$
$C u r r e n t e f f i c i e n c y = \frac{25.67 \times 100}{27} = 95.09$ %

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Similar questions

Q. Electrolysis of a solution of

M n S O_{4}

in aqueous sulphuric acid is a method for the preparation of

M n O_{2}

as the per reaction,

M n^{2 +} (a q .) + 2 H_{2} O \to M n O_{2} + 2 H^{+} (a q .) + H_{2}

Passing a current of

27 A

for

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hours gives one kg of

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Q. Electrolysis of a solution of

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in aqueous sulphuric acid is a method for the preparation of

M n O_{2}

. Passing a current of

27 A

for

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hours gives

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M n O_{2}

. The current efficiency in this process is:

Q. The electrosynthesis of

M n O_{2}

is carried out from a solution of

M n S O_{4}

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(aq). If a current of 25.5 ampere is used with a current efficiency of 85%, how long would it take to produce 1 kg of

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Q. An aqueous solution of NaCl on electrolysis gives

H_{2} (g), C l_{2} (g)

and

N a O H

according to the reaction :

2 C l^{-} (a q) + 2 H_{2} O (l) ⟶ 2 O H^{-} (a q) + H_{2} (g) + C l_{2} (g)

A direct current of 25 amperes with a current efficiency of 62% is passed. Calculate the time required to produce 1 kg

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Q. An aqueous solution of NaCl on electrolysis gives

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C l_{2}

(g) and NaOH according to reaction,

2 C l^{-} (a q) + 2 H_{2} O \to 2 O H^{-} (a q) + H_{2} (g) + C l_{2} (g)

A direct current of 25 ampere with a current efficiency of 62% is passed through 20 litre of NaCl solution (20% by mass) Write down the reactions taking place at the electrodes.
The redox changes are : Cathode:

2 C l^{-} \to C l_{2} + 2 e

Anode :

2 e + 2 H_{2} O \to 2 O H^{-} + H_{2}

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Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method of preparation of MnO2 as per reaction, Mn2+(aq.)+2H2O→MnO2(s)+2H+(aq.)+H2(g)Passing a current of 27 A for 24 hr gives 1 kg of MnO2. What is the value of current efficiency?

The correct option is D 95.1%w=E×i×t96500E(equivalent weight)=Molecular weightchange in oxidation no.E(MnO2)=872=43.5Thus, substituting values1000=43.5×i×24×360096500i=25.67ACurrent efficiency=25.67×10027=95.09%