# How do I balance: MnO4- (aq) + CH3OH (aq) —> Mn2+(aq) + HCO2H (aq)?

I know that H+ and H20 are also a part of the balanced equation, but I have no idea what side to put what on.

MnO4- + 8 H+ + 5e- = Mn2+ + 4 H2O
CH3OH + H2O= HCO2H + H+ + 1e-) x 5

MnO4- + 5 CH3OH + H2O + 3 H+ = Mn2+ + 5 HCO2H

initially, permanganate is an ion, so it is going to be MnO4-. because of this, your first reaction would be as follows, so as that the regularly occurring value is two+ on the two aspects: 5e- + 8H^+ + MnO4^- ? Mn^2+ + 4H2O additionally, your 2d reaction for step #6 is incorrect. There are greater hydrogens that would desire to be accounted for. instead, it is going to be: H2O + CH3OH ? HCO2H + 4H^+ + 4e- Resultantly, you will desire to multiply the 1st reaction with the help of a component of four and the 2d reaction with the help of a component of 5 as a manner to stability the electrons. 20e- + 32H^+ + 4MnO4^- ? 4Mn^2+ + 16H2O 5H2O + 5CH3OH ? 5HCO2H + 20H^+ + 20e- the internet reaction is: 12H^+ + 4MnO4^- + 5CH3OH ? 4Mn^2+ + 5HCO2H + 11 H2O

Also Read :   In a group discussion which option is mostly clearly a clarifying question?

4MnO4^-(aq) + 5CH3OH(aq) + 12H^+(aq)→4Mn^2+(aq) + 5HCO2H(aq) + 11H2O(l)

4MnO4−(aq)+5CH3OH(aq)+12H+(aq)→4Mn2+(aq)+5HCO2H(aq)+11H2O(l)