Chemistry - Electrochemistry Question with Solution | TestHub

$$2{\mathrm{e}}^{-}+{\mathrm{Ca}}^{2+}⟶{\mathrm{Ca}}_{(\mathrm{s})}$$ 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 $=2\times 0.25=0.5\mathrm{F}$