Using the electronic balance method, arrange the coefficients in the reaction equation, the scheme of which is K2MnO4 + HCl (conc,) = MnCl2 + KCl + Cl2 + H2O Determine the oxidizing agent and reducing agent
K2MnO4 + HCl (conc) = MnCl2 + KCl + Cl2 + H2O.
1. Let’s arrange the oxidation states of each element.
K +12Mn +6O -24 + H +1Cl-1 (конц ) = Mn +2Cl -12 + K +1Cl-1 + Cl02 + H +12O-2.
2. Let us write down those elements that have changed their oxidation state.
3.Let’s find the number of electrons that have received or given these elements.
Mn +6 + 4e → Mn +2 │1 oxidizing agent, reduced.
Cl-1 -1e → Cl02 │4 reducing agent, oxidizes.
Chlorine gives up 1 electron, exhibits the properties of a reducing agent, while itself is oxidized.
Manganese takes 4 electrons, an oxidizing agent, and is itself reduced.
4. Find the smallest common multiple for the found number of electrons (for 4 and 1), divide the smallest common multiple by these numbers, respectively. Thus, we will find the coefficients that need to be put into the equation. We start to equalize from the right side of the equation, then from the left. Last but not least, we balance the oxygen.
K2MnO4 +8 HCl (conc) = MnCl2 + 2KCl + 2 Cl2 + 4H2O.