The equation shows a redox reaction between iron(II) chloride and chlorine gas. $$\ce{2FeCl2 + Cl2 -> 2FeCl3}$$ Which equation describes the reduction process in this reaction? \begin{align} \ce{2Cl- &-> Cl2 + 2e-}\tag{A}\\ \ce{Cl2 + 2e- &-> 2Cl- }\tag{B}\\ \ce{Fe^2+ &-> Fe^3+ + e-}\tag{C}\\ \ce{Fe^3+ + e- &-> Fe^2+}\tag{D} \end{align}

I couldn't understand why the answers are not C instead of B.

The correct answer is B. A and D cannot be the answer.

  • 2
    $\begingroup$ Because what is reduction, really? That's when you add electrons to something. $\endgroup$ Sep 30 '15 at 10:12

A redox reaction consists of 2 parts, oxidation and reduction These two parts can be written up as half-equations where one half-equation shows oxidation and the other shows reduction.

It seems that you have mixed up the oxidation and reduction half-equations. Oxidation is losing electrons while reduction is gaining electrons. This can be remembered by the mnemonic:

enter image description here

The question is asking for the reduction half-equation. To do this, you must first recognise what is getting reduced (in other words, what is the oxidant). This can be done by looking at the oxidation states for each element before and after the reaction. Since it is getting reduced, it is gaining electrons and therefore its oxidation state should decrease after the reaction. So by looking at the oxidation states, it should be clear that chlorine is getting reduced as its oxidation state goes from $0$ to $-1$.

Therefore the answer is B

Note: The answer definitely can't be a or c since they are oxidation half-equations

  • 1
    $\begingroup$ Another mnemonic I've found useful is "LEO the plussy cat says GER". Lose electrons for oxidation, (often producing) cations with a plus charge, and gain of electrons is reduction. $\endgroup$
    – Curt F.
    Sep 30 '15 at 20:32

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.