Question

A quantity of electricity equal to $\mathbf{6}\mathbf{.}\mathbf{95}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{C}$passes through an electrolytic cell that contains a solution of$S{n}^{4+}\left(aq\right)$ions. Compute the maximum chemical amount, in moles, ofthat can be deposited at the cathode.

Short answer

The maximum chemical amount in moles, of$Sn\left(s\right)$ that can be deposited at cathode is$0.18mol$

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.

Understanding given parameters

• Quantity of electricity given as$6.95\times {10}^4\mathrm{C}$.
• Solution contains 4 moles of electrons.

Determining charge for 4 moles of electrons

Writing cathode half reaction for the reduction of$S{n}^{4-}$to Sn(s),

$S{n}^{4+}\left(aq\right)+4e^{-}\to Sn\left(s\right)$

Computing maximum chemical amount in moles

Number of moles in a solution formulated as,

$$\begin{gathered} n=\frac{aquantityofelectricity}{chargefor4molesofelectrons} \\ =\frac{6.95\times {10}^{4}C}{3.86\times {10}^{5}Cmo{l}^{-1}} \\ =1.08mol \end{gathered}$$

Therefore, the maximum chemical amount in moles, of Sn (s) that can be deposited at cathode is 1.08 mol.