Calculate the weight of metal deposited when a current of 15 ampere with 75 % current efficiency is passed through the cell for 2 hours. (Electrochemical equivalent of metal = 4 × 10-4)

Question

Calculate the weight of metal deposited when a current of 15 ampere with 75 % current efficiency is passed through the cell for 2 hours. (Electrochemical equivalent of metal = 4 × 10-4) (A) 32.4 g (B) 43.2 g (C) 57.6 g (D) 16.2 g

Tardigrade Admin · Accepted Answer

Applying, W = Z × I × t W =?, Z =4 ×-4; I=15 × (75/100) = (45/4) t =2 hr. =2 × 3600 sec W =4 × 10-4 × (45/4) × 2 × 3600 = 32.4 g

Q. Calculate the weight of metal deposited when a current of $15$ ampere with $75%$ current efficiency is passed through the cell for $2$ hours. (Electrochemical equivalent of metal $= 4 \times 1 0^{- 4}$ )

$32.4 g$

$43.2 g$

$57.6 g$

$16.2 g$

Applying, $W = Z \times I \times t$
$W = ?, Z = 4 \times^{- 4}; I = 15 \times \frac{75}{100} = \frac{45}{4}$
$t = 2 h r$ . $= 2 \times 3600 sec$
$W = 4 \times 1 0^{- 4} \times \frac{45}{4} \times 2 \times 3600 = 32.4 g$

Q. Calculate the weight of metal deposited when a current of 15 ampere with 75% current efficiency is passed through the cell for 2 hours. (Electrochemical equivalent of metal =4×10−4)

32.4g

43.2g

57.6g

16.2g