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In a given reaction like this: $$\ce{FeO4^2- + 3e^- + 8H+ <=> Fe^3+ + 4H2O}$$

From where does the electron come in the acidic medium, as there is no other species to get oxidised in the medium? Also if this were in basic medium then why there is no oxidation of any other species in the medium?

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    $\begingroup$ This is not a reaction, but a half-reaction. The electron comes from the other half-reaction. $\endgroup$ – Ivan Neretin Apr 25 '16 at 10:55
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    $\begingroup$ Expanding on Ivan's comment, what you have here is a reduction half-equation (electrons are being accepted). As the name suggests, this only shows half of the actual reaction. You also need to have the oxidation half-equation (electrons being donated) and add them together to get the entire equation for the reaction. When you add these two half-equations together, the electrons should cancel out. $\endgroup$ – Nanoputian Apr 25 '16 at 11:05
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What you have here is a reduction half-equation as the $\ce{Fe(IV)}$ is accepting electrons to get reduced to $\ce{Fe(III)}$. As the name suggests, this only shows half of the actual reaction as if something is getting reduced, something else must be getting oxidised. So what you need is the oxidation half-equation which shows the species that is getting oxidised and donating electrons. When you add these two half equations the electrons will cancel each other out and you will get your equation for the reaction.

For example, one possible oxidation half-equation is the following: $$\ce{Mg -> Mg^{2+} + 2e-}$$ Now if you times the co-efficients of the reduction half-equation by 2 and times the co-efficients of the oxidation half-equation by 3 and then add them together you will get the following equation: $$\ce{2FeO4^{2-} 3Mg + 16H+ -> 2Fe^{3+} 3Mg^{2+} + 8H2O}$$ As you can see all the electrons balance out and that is your complete equation for the reaction.

Sorry, I don't think I really understand your second question about basic medium. Could you please expand on it to clarify what you mean.

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