Question

Chemical equations of reduction–oxidation (redox) reactions can be quite nontrivial to balance. To do so, you begin with balancing the number of electrons some particles lose in oxidation and other particles gain in reduction. Consider a reaction between potassium permanganateKMnO4 andhydrochloricacidHCl(aq), which can be used in a lab to produce chlorine gas Cl2. In this reaction, a MnO− 4 ion is reduced to a Mn2+ ion by getting electrons from Cl− and losing its oxygen atoms to bind with H+ and form water. Calculate the number of electrons in the processes of reduction, MnO− 4 + H+ + electrons → Mn2+ +H 2O(firsthalf-reaction),and oxidation,Cl−−electrons→Cl2 (second half-reaction). Then multiply those half-reactions with appropriate numerical factors to equate the number of electrons given and taken in these processes. Then add both half-reactions to get the net equation. Finally, add spectator ions considering that the products include potassium and manganese(II) chlorides.

Answer 1

5 electron in reduction and 2 electron in oxidation reaction.

Jx 2 8Ht+5e M Mm04 2+ +4H20 2 c 10 C 16H 2 M 2 Mn0 SC2 t8 420 then eqyat, Add Spectator ims 2kt ond 6 c to boh sides + SCR2 2 K M0 10H C +6 6 C 2 Ma 2K6C t 8 H20 2 KM04 t16HC 2 Mn c2t8 M20 Sc2 +2KC