How many Faraday of electricity is required to deposit $0.8 \mathrm{~g}$ of calcium at cathode by the…
How many Faraday of electricity is required to deposit $0.8 \mathrm{~g}$ of calcium at cathode by the electrolysis of $\mathrm{CaCl}_2$ ?
- $4 \mathrm{~F}$
- $0.04 \mathrm{~F}$
- $2.5 \mathrm{~F}$
- $2 \mathrm{~F}$
Solution
$\begin{aligned}
& \mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_{(\mathrm{s})} \\
& \qquad \mathrm{nf}=2 \\
& \text { moles of calcium }=\frac{0.8}{40}=2 \times 10^{-2} \\
& \text { no. of equivalents of calcium } \\
& =\text { moles } \times \mathrm{nf} \\
& =2 \times 10^{-2} \times 2 \\
& =4 \times 10^{-2}=0.04
\end{aligned}$
To deposit 1 equivalent of electrolyte 1 Faraday electricity is required.
To deposit 0.04 equivalents of electrolyte, $0.04 \mathrm{~F}$ electricity is required.
Asked in: MHT CET 2021 (22 Sep Shift 2)
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