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.


Using Faradays' Law, we have:

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

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


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

$\implies n = 0.049$ 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.

  • $\begingroup$ For one using units will prevent most errors though not in this case. $\endgroup$ – 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 $n = \pu{0.49 mol}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.