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Will $\text{Fe}$ be oxidised to $\text{F}{{\text{e}}^{+2}}$ by reaction with 1 M $\text{HCl}$ ? $\text{E}{}^\circ $ for $\text{Fe/F}{{\text{e}}^{+2}}$=+0.44 volt.

Last updated date: 27th May 2024
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Hint: The oxidation and reduction of any element occur. If the $\vartriangle \text{G}$, (it represents Gibbs free energy and describes the spontaneity of a process) of the reaction is negative for the reaction to take place spontaneously. First find $\vartriangle \text{G}$ of the reaction and then decide.

Complete answer: Let us find the net for the reaction. But we find $\text{E}{}^\circ $, we have first to write the equation dealing $\text{Fe}$ and $\text{HCl}$ in the reactants side to see the final products formed.
$\text{Fe} + 2\text{HCl}\to \text{FeC}{{\text{l}}_{2}} + {{Cl }_{2}}$, this reaction where $\text{HCl}$ is reduced to as the oxidation state has reduced from +2 to 0. But the oxidation state of $\text{Fe}$ which lost 2 electrons is changed from 0 to +2. Let us find $\text{E}{}^\circ $, it is given that $\text{E}{}^\circ $for$\text{Fe/F}{{\text{e}}^{+2}}$=+0.44 volt and $\text{E}{}^\circ $for ${{\text{H}}_{2}}/{{\text{H}}^{+}}$ is 0 volt.
So, $\text{E}{}^\circ $=$\text{E}_{\text{anode}}^{{}^\circ }-\text{E}_{\text{cathode}}^{{}^\circ }$, on anode is $\text{Fe}$ and on cathode is $\text{HCl}$ so the $\text{E}{}^\circ $ will be 0.44-0 volts which is equal to 0.44 volts. $\text{E}{}^\circ $ of the reaction is 0.44 volts.
The relation between $\text{E}{}^\circ $ and $\vartriangle \text{G}$ is $\vartriangle \text{G}$$=\text{-nF}{{\text{E}}^{{}^\circ }}$; where n is the number of electrons involved in the reaction and F is Faraday constant whose value is 96500 C/mol. As,$\text{E}{}^\circ $ is positive so $\vartriangle \text{G}$ will be negative due to – sign present in the formula. $\vartriangle \text{G}$ is negative; which tells us that the reaction will occur spontaneously.
The answer is $\text{Fe}$ will be oxidised to $\text{F}{{\text{e}}^{+2}}$ by the reaction with 1 M $\text{HCl}$.

Note: The question can be solved in two lines only, by just knowing the electrochemical series. We know in electrochemical series oxidation of $\text{Fe}$ lies above of the oxidation of ${{\text{H}}_{2}}$ so, the one which lies above is able to displace the other from its solution. Thus, $\text{Fe}$ gets converted to $\text{F}{{\text{e}}^{+2}}$. So, yes the reaction will proceed forward.

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