(B) The reaction at the cathode is: $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$For $63 \ g$ of $Cu$,the electricity required is $2 \ F$ $(2 \times 96500 \

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(B) The reaction at the cathode is: $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$For $63 \ g$ of $Cu$,the electricity required is $2 \ F$ $(2 \times 96500 \ C)$.So,for $0.4725 \ g$ of $Cu$,the electricity required $(Q)$ is: $Q = \frac{2 \times 96500 \times 0.4725}{63} = 1447.5 \ C$.Using Faraday's law: $Q = I \times t$,where $t = 15 \times 60 \ s = 900 \ s$.$I = \frac{Q}{t} = \frac{1447.5}{900} = 1.608 \ A$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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$CuSO_4$ solution is electrolysed for $15 \ minutes$ to deposit $0.4725 \ g$ of copper at the cathode. The current in amperes required is (Faraday $= 96,500 \ C \ mol^{-1}$,atomic weight of copper $= 63$)

AP EAMCET 2019 All AP EAMCET

A
$0.804$
B
$1.608$
C
$1.206$
D
$0.402$

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Class 12

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Electrochemistry

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Faraday’s law of electrolysis

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$CuSO_4$ solution is electrolysed for $15 \ minutes$ to deposit $0.4725 \ g$ of copper at the cathode. The current in amperes required is (Faraday $= 96,500 \ C \ mol^{-1}$,atomic weight of copper $= 63$)

Similar Questions

Calculate the mass of $Ca$ deposited at the cathode by passing $0.8 \ A$ current through molten $CaCl_2$ for $60 \ minutes$. [Molar mass of $Ca = 40 \ g \ mol^{-1}$] (in $g$)

One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate,$x$ is . . . . . . .

During electrolysis of fused aluminium chloride,$0.9 \ g$ of aluminium was deposited on the cathode. The volume of chlorine liberated at the anode will be .............. $litres$ (at $STP$).

How many coulombs are required for the oxidation of $2 \, mol$ of $H_2O_2$ to $O_2$?

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