The quantity of electricity which deposits 1.08 g of silver from AgNO 3 solution is

Question

The quantity of electricity which deposits 1.08 g of silver from AgNO 3 solution is (A) 96500 coulombs (B) 9650 coulombs (C) 965 coulombs (D) 96.5 coulombs.

Tardigrade Admin · Accepted Answer

Applying Faraday's first law, w=Z Q Z=( text Eq. wt. /96500) Q= quantity of electricity in coulombs w= mass deposited =1.08 g ∴ 1.08=(108/96500) × Q ⇒ Q=(1.08 × 96500/108)=965 coulombs

Q. The quantity of electricity which deposits $1.08 g$ of silver from $A g N O_{3}$ solution is

$96500$ coulombs

$9650$ coulombs

$965$ coulombs

$96.5$ coulombs.

Applying Faraday's first law, $w = ZQ$

$Z = \frac{Eq. wt.}{96500}$

$Q =$ quantity of electricity in coulombs

$w =$ mass deposited $= 1.08 g$

$∴ 1.08 = \frac{108}{96500} \times Q \Rightarrow Q = \frac{1.08 \times 96500}{108} = 965$ coulombs

Q. The quantity of electricity which deposits 1.08g of silver from AgNO3​ solution is

96500 coulombs

9650 coulombs

965 coulombs

96.5 coulombs.