You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Faraday's First Law

An aqueous so...

Question

An aqueous solution of sodium chloride on electroysis gives $H_{2} (g), C l_{2} (g)$ and $N a O H$ according to the reaction:
$2 C l^{-} (a q .) + 2 H_{2} O \to 2 O H^{-} (a q .) + H_{2} (g) + C l_{2} (g)$
A direct current of $25$ ampere with a current efficiency $62$ % is passed through $20 L$ of $N a C l$ solution $(20$ % by mass). Write down the reactions taking place at the anode and cathode. How long will it take to produce $1 k g$ of $C l_{2}$ ? Assume no loss due to evaporation.

Open in App

Solution

Reactions at anode and cathode are:
$2 C l^{-} \to C l_{2} + 2 e^{-}$ (at anode)
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ (at cathode)
$1 k g$ of $C l_{2} = \frac{1000}{71.0} = 14.08 m o l e$
Charge to produce one mole of $C l_{2} = 2 \times 96500 c o u l o m b$
Charge to produce $14.08$ mole of $C l_{2} = 2 \times 96500 \times 14.08 c o u l o m b$
Effective current $= \frac{62}{100} \times 25.0 = 15.5 a m p e r e$
$T i m e = \frac{C h a r g e}{C u r r e n t} = \frac{2 \times 96500 \times 14.08}{15.5}$
$- 175318.7 s e c o n d = 48.699 h o u r$
$O H^{-}$ ions produced $= 2 \times m o l e s o f C l_{2}$
$= 3 \times 14.08 = 28.16 m o l e$
$M o l a r i t y = \frac{M o l e}{V o l u m e} = \frac{28.16}{20} = 1.408 M$ .

Suggest Corrections

Similar questions

Q. An aqueous solution of NaCl on electrolysis gives

H_{2}

(g),

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}

Q. An aqueous solution of NaCl on electrolysis gives

H_{{2 (g)}^{'}}

C l_{2 (g)}

and NaOH according to the reaction

2 C l_{(a q)}^{-} + 2 H_{2} O ⇌ 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 through 20 liters of NaCl solution (20% by weight). Write down the reactions taking place at the anode and the cathode. How long will it take to produce 1 kg of

C l_{2}

? in hrs. (write the value to the nearest integer).

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

C l_{2} (g)

An aqueous solution of NaCl on elctrolysis gives $H_{2} (g)$ , $C l_{2} (g)$ , and $N a O H$ according to the reaction

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

A Direct current (DC) of 25 A with a current efficiency of 62 % is passed through 20 L of NaCl solution (20% by weight). The molarity (M) of the solution with respect to hydroxide ion will be X. The value of X is ( asssume no loss due to evaporation):

(write your answer to nearest integer)

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

24

hours gives one kg of

M n O_{3}

. What is the value of current efficiency? Write the reactions taking place at the cathode and at the anode.

Join BYJU'S Learning Program