MHT CET · Chemistry · Electrochemistry
How many faraday of electricity is required to produce \(10 \mathrm{~g}\) of calcium metal (molar mass \(=40 \mathrm{~g} \mathrm{~mol}^{-1}\) ) from calcium ions?
- A \(1.5 \mathrm{~F}\)
- B \(2.0\mathrm{~F}\)
- C \(0.50 \mathrm{~F}\)
- D \(1.0 \mathrm{~F}\)
Answer & Solution
Correct Answer
(C) \(0.50 \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
\(
2 \mathrm{e}^{-}+\mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_{(\mathrm{s})}
\)
Charge of \(1 \mathrm{~mol}\) electron \(=1\) faraday of electricity To produce \(1 \mathrm{~mol}\) calcium metal \(2 \mathrm{~F}\) electricity is required Moles of calcium \(=\frac{10}{40}=0.25\)
To produce \(0.25 \mathrm{~mol}\) calcium electricity required \(=2 \times 0.25=0.5 \mathrm{~F}\)
2 \mathrm{e}^{-}+\mathrm{Ca}^{2+} \longrightarrow \mathrm{Ca}_{(\mathrm{s})}
\)
Charge of \(1 \mathrm{~mol}\) electron \(=1\) faraday of electricity To produce \(1 \mathrm{~mol}\) calcium metal \(2 \mathrm{~F}\) electricity is required Moles of calcium \(=\frac{10}{40}=0.25\)
To produce \(0.25 \mathrm{~mol}\) calcium electricity required \(=2 \times 0.25=0.5 \mathrm{~F}\)
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