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The redox reaction is

$$\ce{KMnO4-(aq) + 8 H+(aq) + 5 e- -> Mn^2+(aq) + 4 H2O(l)}.$$

Where does $\ce{H+}$ come from in this reaction? What happened to $\ce{K+}$ in potassium permanganate?

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    $\begingroup$ Is this a homework question or an actual research project. Add more details. $\endgroup$
    – M. Farooq
    Mar 17 at 15:32
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    $\begingroup$ Hints: 1) Your redox half-reaction requires acidic conditions. Sulfuric acid is an acid. 2) The potassium ions are spectators, so they are not shown in the redox half-reaction. $\endgroup$
    – Ed V
    Mar 17 at 18:26
  • $\begingroup$ Will the experiment work without adding the acid ? $\endgroup$
    – Tiara
    Mar 17 at 20:09
  • $\begingroup$ Then what would your acid-free redox half-reaction be? $\endgroup$
    – Ed V
    Mar 17 at 21:03
  • $\begingroup$ In the experiment sheet it does not mention anything about sulfuric acid. It only mentions 100ml ground water and 6.5x10-5 mol l-1 potassium permanganate. So in this case how much sulfuric acid is needed? $\endgroup$
    – Tiara
    Mar 18 at 5:12
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In order for you to carry out this reaction, you need potassium permanganate, which provides the ion $\ce{MnO4^-}$. You also need sulfuric acid, and quite a lot of this acid, because the equation requires $\ce{8 H^+}$. And well ! $8$ is much ! So the reaction will not work without acid. And you still need $5$ electrons, which are provided by $\ce{5 Fe^{2+}}$ ion. The $\ce{K+}$ brought by the potassium permanganate remains in the solution as spectator. They don't react.

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  • $\begingroup$ In the experiment sheet it does not mention anything about sulfuric acid. It only mentions 100ml ground water and 6.5x10-5 mol l-1 potassium permanganate. So I’m this case how much sulfuric acid is needed? $\endgroup$
    – Tiara
    Mar 18 at 5:13
  • $\begingroup$ @Lulu. $\pu{100 mL}$ ground water has a meaning. But $\pu{6.5E{-5} mol L^{-1}}$ permanganate has no meaning. One should give you an information like "XXX milliliters of potassium permanganate having a concentration of YYY mole per liter was used for titrating this amount of ground water" $\endgroup$
    – Maurice
    Mar 18 at 8:37

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