MHT CET · Chemistry · Electrochemistry
How many Faraday of electricity is required to deposit \(0.8 \mathrm{~g}\) of calcium at cathode by the electrolysis of \(\mathrm{CaCl}_2\) ?
- A \(4 \mathrm{~F}\)
- B \(0.04 \mathrm{~F}\)
- C \(2.5 \mathrm{~F}\)
- D \(2 \mathrm{~F}\)
Answer & Solution
Correct Answer
(B) \(0.04 \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_{(\mathrm{s})} \\
& \mathrm{nf}=2 \\
& \text {moles of calcium }=\frac{0.8}{40}=2 \times 10^{-2} \\
& \text {no. of equivalents of calcium } \\
& =\text {moles } \times \mathrm{nf} \\
& =2 \times 10^{-2} \times 2 \\
& =4 \times 10^{-2}=0.04
\end{aligned}
\)
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.
\begin{aligned}
& \mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_{(\mathrm{s})} \\
& \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.
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