First slide

Faraday's Laws

Question

When 6 ×10²² electrons are used in the electrolysis of a metallic salt, 1.9 gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So oxidation state of the metal in the salt is

Moderate

A

+ 2
B

+ 3
C

+ 1
D

+ 4

Solution

No. of Coulombs passed $= \frac{6.02 \times 10^{22} \times 96500}{6.02 \times 10^{23}} = 9650$
From Faraday's II law
$m = \frac{Mit}{n \times 96500}; n = \frac{57 \times 9650}{1.9 \times 96500} = 3; Where M = molar mass;$

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