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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:

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.

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.

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  • $\begingroup$ For one using units will prevent most errors though not in this case. $\endgroup$ – A.K. Oct 5 '18 at 4:11
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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 $n = \pu{0.49 mol}$.

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