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A solution of \(\mathrm{Ni}\left(\mathrm{NO}_3\right)_2\) is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of \(\mathrm{Ni}\) is deposited at the cathode?
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Quantity of electricity passed \(=\mathrm{I}, \mathrm{t}\)
\(\begin{aligned}
&=(5 \mathrm{~A}) \times(20 \times 60 \mathrm{sec})=6000 \mathrm{C} \\
&\mathrm{Ni}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}
\end{aligned}\)
Thus, \(2 \mathrm{~F}\), i.e., \(2 \times 96487 \mathrm{C}\) of charge deposit \(=1\) mole of \(\mathrm{Ni}=58.7 \mathrm{~g}\)
\(\therefore 6000 \mathrm{C}\) of charge will deposite \(=\frac{58.7 \times 6000}{2 \times 96487}=1.825 \mathrm{~g}\) of \(\mathrm{Ni}\).
\(\begin{aligned}
&=(5 \mathrm{~A}) \times(20 \times 60 \mathrm{sec})=6000 \mathrm{C} \\
&\mathrm{Ni}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}
\end{aligned}\)
Thus, \(2 \mathrm{~F}\), i.e., \(2 \times 96487 \mathrm{C}\) of charge deposit \(=1\) mole of \(\mathrm{Ni}=58.7 \mathrm{~g}\)
\(\therefore 6000 \mathrm{C}\) of charge will deposite \(=\frac{58.7 \times 6000}{2 \times 96487}=1.825 \mathrm{~g}\) of \(\mathrm{Ni}\).
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