Question

A quantity of electricity equal to ${\mathbf{\text{9.263×10}}}^{\mathbf{\text{4}}}\mathbf{\text{C}}$passes through a galvanic cell that has an Ni(s) anode. Compute the maximum chemical amount, in moles, of${\mathbf{\text{Ni}}}^{\mathbf{\text{2+}}}\mathbf{\text{(aq)}}$that can be released into solution.

Short answer

The maximum chemical amount in moles, of $N{i}^{2+}\left(aq\right)$that can be released into solution is $0.48mol$.

Step-by-step solution

Using faraday constant in electrolysis calculations

Electricity is the flow of electrons. To calculate the charge for 1 mole of electron, we need to know how to relate number of moles of electrons which flow to the measured quantity of electricity.

$n=\frac{aquantityofelectricity}{ch\arg efor\text{2}molesofelectrons}$

$1faraday=9.65\times {10}^{4}Cmo{l}^{-1}$

Understanding given parameters

• Quantity of electricity given as$9.263\times {10}^{4}C$.

• Solution contains 2 moles of electrons.

Determining charge for 4 moles of electrons

Writing anode half reaction for the oxidation of $Ni\left(s\right)$to $N{i}^{2+}\left(aq\right)$,

$N{i}_{-}\left(s\right)\to N{i}^{2+}\left(aq\right)+2e^{-}$

$\begin{array}{c}Ch\arg efor2molesofelectrons=2\times 9.65\times {10}^{4}Cmo{l}^{-1}\\ =1.93\times {10}^{5}Cmo{l}^{-1}\end{array}$

Computing maximum chemical amount in moles

Number of moles in a solution formulated as,

$\begin{array}{c}n=\frac{aquantityofelectricity}{ch\arg efor\text{2}molesofelectrons}\\ =\frac{9.65\times {10}^{4}C}{1.93\times {10}^{5}Cmo{l}^{-1}}\\ =0.48mol\end{array}$

Therefore, the maximum chemical amount in moles, of$N{i}^{2+}\left(aq\right)$ that can be released into solution is $0.48mol$.