# Determining iron content by titration

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?

• Is this a homework question or an actual research project. Add more details. Mar 17 at 15:32
• 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.
– Ed V
Mar 17 at 18:26
• Will the experiment work without adding the acid ? Mar 17 at 20:09
• Then what would your acid-free redox half-reaction be?
– Ed V
Mar 17 at 21:03
• 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? Mar 18 at 5:12

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.
• @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" Mar 18 at 8:37