Question

When 9.65 coulombs of electricity is passed through a solution of silver nitrate

(atomic weight of taking as 108) the amount of silver deposited is

[EAMCET 1992; KCET 2000]

• A. 10.8 mg
• B. 5.4 mg
• C. 16.2 mg
• D. 21.2 mg

Answer 1

Answer: (A)

Step 1:

According to Faraday's Law of Electrolysis:

$W=\frac{I\times t\times M}{nF}$

where, W = wight deposited

I = current

F = Faraday's constant 96500

t = time

M = Molar mass

n = number of electrons

Given that :

$It=9.65coulomb$

Molar mass of Silver, $M=108$ g/mol

n = 1

Step 2:

$W=\frac{9.65\times 108}{1\times 96500}$

$W=0.0108g$

$W=10.8mg$

Hence, option (A) is correct.