In producing chlorine through electrolysis, 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (ECE of chlorine is 0.367 × 10-6 kg C-1)

Question

In producing chlorine through electrolysis, 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (ECE of chlorine is 0.367 × 10-6 kg C-1) (A) 17.6 × 10-3 kg (B) 16.6 × 10-3 kg (C) 15.6 × 10-3 kg (D) 14.6 × 10-3 kg

Tardigrade Admin · Accepted Answer

As P=VI ∴ I=(P/V)=(100 ×103 W/125 V) =800 A According to Faraday's first law of electrolysis m=ZIt Here, Z=0.367 × 10-6 kg C-1 I=800 A t=1 min=60 s ∴ m=(0.367 × 10-6 kg C-1)(800 A)(60 s) =17.6×10-3 kg

Q. In producing chlorine through electrolysis, $100 kW$ power at $125 V$ is being consumed. How much chlorine per minute is liberated?
$(ECE$ of chlorine is $0.367 \times 1 0^{- 6} k g C^{- 1})$

$17.6 \times 1 0^{- 3} k g$

$16.6 \times 1 0^{- 3} k g$

$15.6 \times 1 0^{- 3} k g$

$14.6 \times 1 0^{- 3} k g$

Q. In producing chlorine through electrolysis, 100kW power at 125V is being consumed. How much chlorine per minute is liberated? (ECE of chlorine is 0.367×10−6kgC−1)

17.6×10−3kg

16.6×10−3kg

15.6×10−3kg

14.6×10−3kg

As P=VI ∴I=VP​=125V100×103W​ =800A According to Faraday's first law of electrolysis m=ZIt Here, Z=0.367×10−6kgC−1 I=800A t=1min=60s ∴m=(0.367×10−6kgC−1)(800A)(60s) =17.6×10−3kg

Q. In producing chlorine through electrolysis, $100 kW$ power at $125 V$ is being consumed. How much chlorine per minute is liberated?
$(ECE$ of chlorine is $0.367 \times 1 0^{- 6} k g C^{- 1})$

$17.6 \times 1 0^{- 3} k g$

$16.6 \times 1 0^{- 3} k g$

$15.6 \times 1 0^{- 3} k g$

$14.6 \times 1 0^{- 3} k g$