Question

How much quantity of electricity in coulomb is required to deposit 1.346 × 10^-3 kg of Ag in 3.5 minutes from AgNO₃ solution?
( Given: Molar mass of Ag is 108 × 10^-3 kg mol^-1 )

Answer 1

Given:
Mass of Ag deposited = 1.346 x 10^-3 kg
Time (t) = 3.5 min = 3.5 x 60s
Molar mass of Ag = 108 x 10^-3 kg mol^-1

To find: 
Quantity of electricity required in coulomb:

Formulae:
1. Mole ratio = `"Moles of product formed in half reaction"/"Moles of electrons required in half reaction"`

2. Mass of the substance produced = `["I(A)" xx"t(s)"]/[96500"(C/mol e-)" xx "Mole ratio" xx "Molar mass of substance"]`

3. Quantity of electricity in coulomb (Q) = I ( in amp ) x t( in sec )

Calculation:
The half reaction for the formation of Ag is,
\[\ce{ Ag_(aq)^+ + e^- -> Ag_(s)}\]

From formula (1),
Mole ratio = `"Moles of Ag"/"Moles of electrons"`

= `[1( "mol Ag" )]/[1 ("mol" e^-)]`

= 1 mol Ag/mol e^-

From formula (2),
1.346 x 10^-3 
= `(I xx 3.5 xx 60)/( 96500 ) xx 1 xx 108 xx 10^-3`

∴ I = `( 1.346 xx 10-3 xx 96500)/(3.5 xx 60 xx 108 xx 10^-3)`

= `[ 129889 xx 10^-3]/[22680 xx 10^-3]`

= 5.727 A

From formula (3),
Q = It = 5.727 x 3.5 x 60 = 1202.67 C
∴ The quantity of electricity required is 1202.67 C.