I am a total noob in chemistry, so sorry in advance if this is a very stupid question... I am trying to solve this redox reaction:

$$\ce{KClO3 + FeSO4 + H2SO4 -> Fe2(SO4)3 + KCl}$$ $$\ce{K^{+1}Cl^{+5}O^{-6}3 + Fe^{+2}S^{+6}O^{-8}_4 + H^{+2}2S^{+6}O^{-8}4 -> Fe^{+6}2(S^{+6}O^{-8}4)3 + K^{+1}Cl^{-1}}$$

When I analyze this I can see the following things:

  • Fe lost some electrons, it went from +2 to +6
  • Cl went from +1 to -1: a gain of elecetrons


  • I see something happened to the SO ions as well: on the left side you have two times $S^{+6}O^{-8}_4$ and on the right three times so something changed here as well. What happened? Do I have to take this into consideration as well for the calculation? And on top of that $O_3$ on the left side disappeared.

My current guess is that you never have to look at the change in oxidation state of oxygen atoms. Is that correct?

  • 2
    $\begingroup$ First of all, when talking about oxidation states, normally you don't multiply. You just write $\ce{Fe2^3+}$, and people will get it (correctly) as two times $\ce{Fe^3+}$ with a total "charge" of +6; you don't have to actually write +6 to achieve that. Also, pay attention to the formatting, especially the \ce{} thing. As for the $\ce{SO4^2-}$ ion, nothing happens to it. $\endgroup$ – Ivan Neretin Aug 2 '16 at 14:01
  • $\begingroup$ @IvanNeretin I thought smth did happen to $SO^{2-}_4$ as you have three of them on the right side and only two on the left side. $\endgroup$ – privetDruzia Aug 2 '16 at 14:40
  • $\begingroup$ No, it doesn't work like that. Until the reaction is balanced, you have no idea how much of anything do you have. And once it is balanced, you'll have as much on the right as on the left. $\endgroup$ – Ivan Neretin Aug 2 '16 at 15:03
  • $\begingroup$ yes you must balance the equation, meaning that there are the same number of atoms of each type on both sides . The ions present are Fe$^{2+}$, Fe$^{3+}$, ClO$_3^-$, SO$_4^{2-}$, K$^+$, Cl$^-$ and H$^+$ $\endgroup$ – porphyrin Aug 2 '16 at 16:07

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