Redox reaction: Iron(III) and Iodide ion I'm a real beginner in chemistry and I'm currently reading about redox reactions. I'm quite confused about this simple redox reaction:


  ${\text{Iron(III) ion + Iodide ion} \rightleftharpoons \text {Iron(II) ion + Iodine}}$
  \begin{array}{|c|c|c|c|c|c|}
\hline
\text{Oxidation numbers}&amp; +3 &amp; -1 &amp;\rightleftharpoons &amp; -1 &amp; 0\\ \hline
 &amp; Fe^{3+}&amp; I^-&amp;\rightleftharpoons &amp; Fe^{2+}&amp;I_2\\ \hline
\end{array}


I understand the principles of transferring electrons. But how would I come from the upper formula to the result that the Iodine exists two times on the right side? After reading the reaction formula I would have written the Iodine just once (not as seen here twice).
Iodine exists as a diatomic molecule. This reaction indicates how the iodide ion is converted into diiodine. Your reaction just indicates the electron transfer: you still need to balance the reaction.
You need an $\ce{I_2}$ molecule on the right hand side, which would need two iodide ions on the left hand side:
$$\ce{2 I- -&gt; I_2 + 2e-}$$
To accept two electrons, we would need two $\ce{Fe^3+}$ ions:
$$\ce{2 Fe^3+ + 2 e- -&gt; 2 Fe^2+}$$
Now we have a balanced reaction. All the electrons donated by iodine are accepted by iron. Adding up the two reactions and cancelling the electrons gives us:
$$\ce{2 Fe^3+ + 2I- -&gt; 2 Fe^2+ + I_2}$$