Question

MnO₄^-+ S^2- ? S + MnO₂          Note: S is elemental

Indicate the numeric coefficient in front of the following compounds, after balancing the reaction above

MnO₄^-

S^2-

S

MnO₂  

Answer 1

This a redox reaction, (reduction-oxidation) that means, there is electron (charges) transfer. It cannot be balnaced in the normal way. In order to balnce it you mus write the half reaction separatelly.

The half reaction only has the compund that reduces (gain electrons) or oxidizes (loses electrons), How do I know whom reduces or oxidixes? you must check the oxidation state, see:

MnO₄^- -----> MnO₂
Oxygen have always (not always but must cases) a charge of -2 so, in MnO₄^- the total charge is -1 and there are 4 oxygens with -2, we calculate the chrge of Manganese then (x):

Where x is the Mn charge, 4(-2) is the charge of all oxygen altogether and -1 the total ion charge. Calculate x:

So the charge of Mn is 7.

Now the same with MnO_2. Total charge is 0, charge of oxygen is -2 and there are 2 oxygens so is -2*2 = -4:

Charge of Mn is 4. Initially it was +7 and then after the reaction +4, it became less positive, it means it gained electrons (which have negstive charge) that means it reduced. If something reduces something MUST oxidize and that is S^2-. It pass from S^2- to S (loses charge) it became less negative, i loses electrons.

Now write the half equations and balance it:

MnO₄^- -----> MnO₂

There is one Mn in both side but different amounts of Oxygens, oxygens are balanced with water, there are twi oxygens missing in the right side so you must write 2 waters:
MnO₄^- -----> MnO₂ + 2H₂O

No we got Hydrogens that need to be balanced, we put H⁺ to balance it,
MnO₄^- + 4H⁺  -----> MnO₂ + 2H₂O

Finally we must balance the charges. In the left side we gat 1 negative charge and 4 positive: -1 + 4 = +3
and in the right side 0. we put electrons to make both side equals:
MnO₄^- + 4H⁺ + 3e^- -----> MnO₂ + 2H₂O

Now left side is 0 charge.

we do the same with the other half reaction:
S^2- ----> S + 2e^-

You can see I balanced the charges with two electrons. Finally we write the halfs reaction as a complete one making an addition:

MnO₄^- + 4H⁺ + 3e^- -----> MnO₂ + 2H₂O
S^2- ------> S + 2e^-

The addition is like a normal mathematical equation addition: equal stuff can add and substract:
MnO₄^- + S^2- +4H⁺ + 3e^- -----> MnO₂ + S + 2H₂O + 2e^-

As you can seethere are 3 electrons in the left side and two in the right, they cancel each other bacause is like a mathematical equation:

MnO₄^- + S^2- +4H⁺ + 1e^- -----> MnO₂ + S + 2H₂O