# Faraday's Law of Electrolysis problem

Question:

How many moles of iron will by passage of $$\pu{4 A}$$ of current through $$\pu{1 L}$$ of $$\pu{0.1 M}$$ $$\ce{Fe^{3+}}$$ solution for $$1$$ hour. Assume $$\ce{Fe^{2+}}$$ is not present in the solution initially.

Attempt:

$$\text{mass delivered = } \mathrm {Z\times I\times t} = n\times M$$ where $$\mathrm Z = \dfrac{\text{Equivalent mass }}{\pu{1F}}$$

Equivalent mass of Iron here would be $$\dfrac{M}{3}$$ where M is molar mass.

So,

$$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}$$

$$\implies n = 0.049$$ moles.

But answer given is $$\pu{0.0245 moles}$$

Where have I gone wrong? I can't figure out why there's a difference of factor of 2 in the answer.

• For one using units will prevent most errors though not in this case. – A.K. Oct 5 '18 at 4:11

I think the key part of the answer is

Assume $$\ce{Fe^{2+}}$$ is not present in the solution $$\color{red}{\underline{\text{initially}}}$$.

Since iron reacts with $$\ce{Fe^3+}$$ to form $$\ce{Fe^2+}$$ via:

$$\ce{Fe^0 + 2 Fe^3+ -> 3 Fe^2+}\tag 1$$

You have to account for this additional reaction. If the reaction did not produce any $$\ce{Fe^2+}$$ as you assumed then of the $$\pu{0.100 mol}$$ of $$\ce{Fe^3+}$$ in the solution then $$\pu{0.051 mol}$$ would remain after and you would be correct, but it does react. Now, my math says that the answer is more like

$$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}\tag 2$$

$$\implies n = \pu{0.0497 mol} = \pu{0.050 mol}$$

Which means $$\pu{0.050 mol}$$ of $$\ce{Fe^3+}$$ remains which must be neutralized by the $$\pu{0.050 mol }\ce{Fe^0}$$. From equation 1 we can see that it takes one $$\ce{Fe^0}$$ for every 2 $$\ce{Fe^3+}$$ which menas $$\pu{0.025 mol}$$ of $$\ce{Fe^0}$$ is needed to react with $$\pu{0.050 mol } \ce{Fe^3+}$$ and your remaining $$\ce{Fe^0}$$ is $$\pu{0.025 mol}$$ or $$\pu{0.0245 mol}$$ in the case of $$n = \pu{0.49 mol}$$.