Asked by Addam

Using this redox equation:
Mn(OH)2(s) + MnO4-(aq) → MnO42-(aq)

When the equation is balanced with smallest whole number coefficients, what is the coefficient for OH-(aq) and on which side is it present?

------I think its 4 and on the reactant side but i don't know
Answered by DrBob222
It's on the reactant side but it isn't 4. If you'll post your half cell reactions I'll find the error.
Answered by Addam
3e- + 2H2O + Mn(OH)2(s) → MnO4^(2-)(aq) +6H + 4e-

1e- + MnO4-(aq) → MnO4^(2-)(aq)

----this is what I got after balancing the charges with e- but this maybe completely wrong, I took a guess.

The choices given are:
4, reactant side
4, product side
6, reactant side
6, product side
Answered by DrBob222
Adam, please see the responses to Lucy, John Butch, etc.
Your mistake here is that you are balancing the reaction in acidic solution. MnO4^- goes to MnO4^2- in basic solution. You seem to be OK with these so I'll leave that with you but if you have trouble, please re-post and I can help you through it.
Answered by DrBob222
This info isn't made obvious in most chem classes and isn't prominent in texts so here is a tid bit of info that may prove useful to you.
permanganate (MnO4)^- goes to manganate (MnO4)^2-in basic solution, to MnO2 is neutral or near neutral solution and to Mn^2+ in acidic solution. My experience is that it must be highly basic to get manganate; otherwise you get MnO2, that brown ooky looking stuff that's unmistakably MnO2.
Answered by Anon
I'm pretty sure the answer is "4, reactant side"
Answered by Anonymous
3Mn(OH)2 + 2MnO4^(-) --> 5MnO2 + 2H2O + 2OH^(-)
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