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

Faraday's First Law

The amount of...

Question

The amount of chlorine evolved by passing $2$ A of current in an aqueous solution of $N a C l$ for $30$ minutes is:

2.64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.32

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.62

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.22

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $1.32$ g
Consider the reaction, $N a C l \to N a^{+} + C l^{-}$
$2 C l^{-} + 2 e^{-} \to C l_{2}$
We know that,
$Q = I \times t = 2 \times 30 \times 60 = 3600$ Coulombs
As, $2 \times 96500$ Cooloumbs produce = $71$ g of $C l_{2}$
So, $3600$ Coulombs will produce $= \frac{71}{2 \times 86500} \times 3600$
$= 1.32 g$ of $C l_{2}$

Suggest Corrections

Similar questions

2

30

N a C l

20

6

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

N a O H

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

C l_{2} (g)

H_{2}

C l_{2}

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

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

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

Join BYJU'S Learning Program