Question

find the mass of sodium metal deposited at cathode when 0.1 ampere of current is passed for half hour and the process has 75% efficiency?

​

Answer 1

Answer:

Explanation:

from faraday's first law of electrolysis

W = ZIt

=(E/96500) × 0.1 × 30 × 60

= 0.043 gm

again,

efficiency = (actual mass/theoritical mass)×100%

(0.043×75)/100= actual mass

= 0.032 gm

Answer 2

We have to find the mass of sodium metal deposited at cathode when 0.1 ampere of current is passed for half hour and the process has 75 % efficiency.

solution : current = 0.1 Ampere, time = 30 min = 30 × 60 = 1800 sec and efficiency = 75 %

so charge = efficiency × current × time

= 0.75 × 0.1 × 1800 C = 135C

we know, 1 F = 96500C charge liberates on mole of electrons.

so, no of moles of electrons = charge/96500C

= 135/96500

= 0.0014 mol

we know

at Cathode,

Na⁺ + e¯ => Na

one mole of sodium is deposited by one mole of electrons.

so, 0.0014 mol of sodium will be deposited by 0.0014 mop of electrons.

so mass of sodium = no of moles × atomic mass of sodium

= 0.0014 × 23

= 0.0322g

Therefore 0.0322 g of sodium metal deposited at cathode.