A solution of copper sulphate is electrolysed | vedclass

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(A) The mass of the substance deposited during electrolysis is given by Faraday's law: $m = (I 	imes t 	imes M) / (n 	imes F)$.Here, the current $I

Vedclass · Accepted Answer

$(1)$ $0.2938 	ext{ g}$

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(A) The mass of the substance deposited during electrolysis is given by Faraday's law: $m = (I 	imes t 	imes M) / (n 	imes F)$.Here, the current $I = 1.5 	ext{ A}$, time $t = 10 	ext{ minutes} = 600 	ext{ s}$, molar mass $M = 63 	ext{ g mol}^{-1}$, and $n = 2$ (since $Cu^{2+} + 2e^{-} ightarrow Cu$).Substituting the values: $m = (1.5 	imes 600 	imes 63) / (2 	imes 96487)$.$m = 56700 / 192974 \approx 0.2938 	ext{ g}$.Thus, the mass of copper deposited is $0.2938 	ext{ g}$.

$A$ solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at the cathode is:
(Given: Molar mass of $Cu = 63 \text{ g mol}^{-1}$; $1F = 96487 \text{ C mol}^{-1}$)

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$A$ solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at the cathode is:(Given: Molar mass of $Cu = 63 \text{ g mol}^{-1}$; $1F = 96487 \text{ C mol}^{-1}$)