# How to apply Faraday's law to calculate the charge of metal ions?

Electrolysis of a molten ionic compound with a current of $\pu{0.50 A}$ for $\pu{30 min}$ yielded $\pu{0.700 g}$ of a metallic element at the cathode. If the element has a relative atomic mass of $150.0$, calculate the charge on the metal ions.

My attempt at solving: $$n = \frac{m}{M} = \frac{0.7}{150} = \pu{4.67e-3}\\ Q = Ft = 0.5 \times (30 \times 60) = \pu{900 C}$$

The main problem I'm having here is that, how do I relate this unknown metallic ionic compound to the amount of electrons gained in this reaction?

After I obtain that can I just use $$Q = n\ce{e-} \times F?$$

I then solved for $$F = Q/n = \pu{1.92E5 F}$$. I was experimenting with some stuff and divided this by $\pu{96500 C}$ which got me $1.99$. The correct answer was $\pu{2 F}$.

Is this just coincidence because I don't see the connection between dividing by $\pu{1 F} = \pu{96500 C}$ when I already solved for $F$?

• Use the formula given by @ Kai, but use SI units which means putting time into seconds to find $Q = It$ and $F = 96485.3 \pu{ C mol^{-1}}$. The charge z comes out as 2 to the nearest whole number. Feb 23 '17 at 17:10

When you're confused by a problem like this it is definitely time to use dimensional analysis.

For the amount of electrons $\text{mol}_e$

\begin{align} \text{The ampere} &= \text{Amp} = 1.00 \frac{\text{coulomb}}{\text{second}} \approx 6.242\times10^{18}~\frac{\text{electrons}}{\text{second}}\\ \text{Farady} &= 26.801 \frac{\text{Amp}\times\text{hours}}{\text{mol}_e}\\ 30~\text{minutes} &= \frac{30 \text{minutes}}{60~\text{minutes/hour}} = 0.500~\text{hours}\\ \text{mol}_e &= \frac{0.50~\text{Amp} \times 0.500~\text{hours}}{26.801~\text{Amp}\times\text{hours}\text{/mol}_e} = 9.328\times10^{-3}~\text{mol}_e \end{align}

For the amount of substance of metal $\text{m}_m$

$$\text{mol}_m = \frac{0.700~\text{g}}{150~\text{g/mol}_m} = 4.667\times10^{-3}~\text{mol}_m$$

Charge per atom = $z$

$$z = \frac{\text{mol}_e}{\text{mol}_m}= \frac{9.328\times10^{-3}}{4.667\times10^{-3}} = 1.998 = 2.0 \frac~{\text{electrons}}{\text{atom}}$$

$$m=\left(\frac{Q}{F}\right)\left(\frac{M}{z}\right)$$

In the given question, we have to find $z$ which is the valency of the metallic ions.
Use $Q=IT$ (in Ampere-Hours), $F=26.8$ Ampere hour per equivalent.