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A metal plate of dimension of $\left(1 \times 2 \mathrm{~cm}^2\right)$ has to be coated on both the sides by $\mathrm{Cu}$ metal. How long does it take to deposit $\mathrm{Cu}$ of $0.01 \mathrm{~cm}$ thickness, if $1.5 \mathrm{~A}$ current is used? [Electrochemical equivalence of $\mathrm{Cu}$ is $0.0003 \mathrm{~g} / \mathrm{C}$ and the density of $\mathrm{Cu}$ is $\left.9 \mathrm{~g} / \mathrm{cm}^3\right]$
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1299 Upvotes
Verified Answer
The correct answer is:
800 s
Step I : Calculate volume
$$
\begin{aligned}
& =\text { area } \times \text { thickness }(\text { two times) } \\
& =1 \times 2 \mathrm{~cm}^2 \times 2 \times 0.01=0.04
\end{aligned}
$$
Step II : Mass of material deposited
$$
=0.04 \times 9=0.36 \mathrm{~g}=\text { volume } \times \text { density }
$$
Step III : According to Faraday law,
$$
\begin{aligned}
m & =Z i t \\
i & =1.5 \mathrm{~A} \\
Z & =0.0003 \mathrm{~g} / \mathrm{C} \\
t & =\frac{m}{Z i}=\frac{0.36}{0.0003 \times 1.5} \\
t & =800 \mathrm{~s}
\end{aligned}
$$
$$
\begin{aligned}
& =\text { area } \times \text { thickness }(\text { two times) } \\
& =1 \times 2 \mathrm{~cm}^2 \times 2 \times 0.01=0.04
\end{aligned}
$$
Step II : Mass of material deposited
$$
=0.04 \times 9=0.36 \mathrm{~g}=\text { volume } \times \text { density }
$$
Step III : According to Faraday law,
$$
\begin{aligned}
m & =Z i t \\
i & =1.5 \mathrm{~A} \\
Z & =0.0003 \mathrm{~g} / \mathrm{C} \\
t & =\frac{m}{Z i}=\frac{0.36}{0.0003 \times 1.5} \\
t & =800 \mathrm{~s}
\end{aligned}
$$
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