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
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
3Mn(OH)2 + 2MnO4^(-) --> 5MnO2 + 2H2O + 2OH^(-)
It's on the reactant side but it isn't 4. If you'll post your half cell reactions I'll find the error.
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.
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.
I'm pretty sure the answer is "4, reactant side"
To determine the coefficient for OH-(aq) in the given redox equation and on which side it is present, we need to balance the equation.
1. Start by counting the number of atoms on each side of the equation.
On the left side (reactant side), we have:
- 1 Mn atom from Mn(OH)2
- 2 O atoms from Mn(OH)2
- 1 H atom from Mn(OH)2
On the right side (product side), we have:
- 1 Mn atom from MnO42-
- 4 O atoms from MnO42-
2. Since O atoms are not balanced, let's balance them first.
We can achieve this by adding water (H2O) molecules to the appropriate side of the equation:
Mn(OH)2(s) + MnO4-(aq) → MnO42-(aq) + H2O
Now, the oxygen atoms are balanced:
On the left side, we have 2 O atoms from Mn(OH)2, and on the right side, we have 4 O atoms from MnO42-.
3. Next, let's balance the hydrogen atoms.
Since there are no hydrogen atoms on the right side, we can balance this by adding hydroxide ions (OH-) to the left side:
Mn(OH)2(s) + MnO4-(aq) + 4 OH-(aq) → MnO42-(aq) + H2O
Now, the hydrogen atoms are balanced:
On the left side, we have 2 H atoms from Mn(OH)2, and on the right side, we have 2 H atoms from H2O.
The coefficient for OH-(aq) is indeed 4, and it is present on the left side (reactant side) of the balanced equation:
Mn(OH)2(s) + MnO4-(aq) + 4 OH-(aq) → MnO42-(aq) + H2O