(A) The volume of $Cu$ deposited is $V = \text{Area} \times \text{thickness} = (20 \times 5) \ cm^2 \times (3.6 \times 10^{-4} \ cm) = 0.036 \ cm^3$.

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(A) The volume of $Cu$ deposited is $V = \text{Area} \times \text{thickness} = (20 \times 5) \ cm^2 \times (3.6 \times 10^{-4} \ cm) = 0.036 \ cm^3$. The mass of $Cu$ deposited is $W = V \times d = 0.036 \ cm^3 \times 8.9 \ g/cm^3 = 0.3204 \ g$. Using Faraday's law,$W = \frac{M \times Q}{n \times F}$,where $M = 63.5 \ g/mol$,$n = 2$,and $F = 96500 \ C/mol$. $Q = \frac{W \times n \times F}{M} = \frac{0.3204 \times 2 \times 96500}{63.5} \approx 973.81 \ C \approx 974 \ C$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Medium

Copper is to be electrodeposited on a nickel block of $(20 \times 5) \ cm^2$ area by using $CuSO_4$ as electrolyte. How much quantity of electricity is needed to deposit a $3.6 \ \mu m$ layer of copper (in $C$)? [Atomic weight of $Cu = 63.5 \ g \ mol^{-1}$,Density of $Cu = 8.9 \ g/cm^3$]

TS EAMCET 2020 All TS EAMCET

A
$974$
B
$580$
C
$1080$
D
$365$

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Electrochemistry

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Faraday’s law of electrolysis

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Copper is to be electrodeposited on a nickel block of $(20 \times 5) \ cm^2$ area by using $CuSO_4$ as electrolyte. How much quantity of electricity is needed to deposit a $3.6 \ \mu m$ layer of copper (in $C$)? [Atomic weight of $Cu = 63.5 \ g \ mol^{-1}$,Density of $Cu = 8.9 \ g/cm^3$]

Similar Questions

If a current of $5 \ A$ is passed through a zinc sulfate solution for $40 \ \text{minutes}$,what is the mass of zinc deposited at the cathode (in $g$)?

In acidic medium,$MnO_4^-$ is converted to $Mn^{2+}$ when it acts as an oxidant. The quantity of electricity required to reduce $0.06 \ mol$ of $MnO_4^-$ would be ........... $F$.

$Al_2O_3$ is reduced by electrolysis at low potential and high current. If a current of $4 \times 10^4 \ A$ is passed through molten $Al_2O_3$ for $6 \ h$,how many grams of $Al$ will be produced? (Assume $100\%$ current efficiency and atomic mass of $Al = 27 \ g \ mol^{-1}$)

For the given reaction how much quantity of electricity in Coulomb is required? $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$

When $9.65 \ C$ of electricity is passed through a solution of silver nitrate (atomic weight of $Ag = 108$),the amount of silver deposited is $............ \ mg$.

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