Question

For how much time, $10$ ampere electric current should be passed through a dilute aqueous $Ni{SO}_4$ solution during electrolysis using inert electrode, in order to get $5.85g$ of Nickel?

[At. weight of $Ni=58.5gm$]

• A. $965sec$
• B. $3860sec$
• C. $1930sec$
• D. $9650sec$

Answer 1

Answer: (C)

$1930sec$

$\text{Mass of Ni}=\frac{\text{I(A)}\times \text{t(s)}}{{\text{96500 C/mol e}}^{-}}\times \text{mole ratio}\times \text{molar mass of Ni}$

$\text{5.85 g}=\frac{\text{10 A}\times \text{t(s)}}{{\text{96500 C/mol e}}^{-}}\times \frac{\text{1 mol Ni}}{{\text{2 mol e}}^{-}}\times \text{58.5 g/mol}$

$t\left(s\right)=\text{1930 sec}$