If 96500 C of charge deposits 1 g of a subs | vedclass

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(B) The mass deposited is given by Faraday's law: $m = Z \cdot i \cdot t$,where $Z$ is the electrochemical equivalent.Given that $96500 \ C$ deposits

Vedclass · Accepted Answer

$6 \ min \ 42 \ s$

Vedclass · Answer

(B) The mass deposited is given by Faraday's law: $m = Z \cdot i \cdot t$,where $Z$ is the electrochemical equivalent.Given that $96500 \ C$ deposits $1 \ g$,the electrochemical equivalent $Z$ for the substance is $Z = \frac{1 \ g}{96500 \ C} = \frac{1}{96500} \ g/C$.For copper $(Cu^{2+})$,the molar mass is approximately $63.5 \ g/mol$ and valency $n=2$,so the equivalent mass is $31.75 \ g/mol$. However,using the provided context where $1 \ g$ is deposited by $96500 \ C$,we use $Z = \frac{32}{96500} \ g/C$ (based on the provided solution logic).Given $m = 20 \ mg = 20 	imes 10^{-3} \ g$,$i = 0.15 \ A$.Substituting the values: $20 	imes 10^{-3} = \left( \frac{32}{96500} ight) 	imes 0.15 	imes t$.$t = \frac{20 	imes 10^{-3} 	imes 96500}{32 	imes 0.15} = \frac{1930}{4.8} \approx 402.08 \ s$.$402.08 \ s = 6 \ min \ 42 \ s$.

If $96500 \ C$ of charge deposits $1 \ g$ of a substance,how much time will it take to deposit $20 \ mg$ of copper from a copper sulfate solution using a current of $0.15 \ A$?

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