Can't figure this one out....how do you balance MnO4- + SO3-2 ====> SO4-2 + Mn+2?
Mn is +7 on the left; +2 on the right.
S is +4 on the left;6 on the right.
Here is one I did for Nessy that gifes the method I use.
http://www.jiskha.com/display.cgi?id=1274415665
To balance the chemical equation MnO4- + SO3-2 ===> SO4-2 + Mn+2, you can follow a step-by-step process known as balancing by the method of half-reactions. Here's how to do it:
Step 1: Assign oxidation numbers to each element in the equation. The oxidation number is a way to keep track of electron distribution in a compound. In this equation:
MnO4-: The overall charge of the permanganate ion is -1. Since oxygen is typically assigned an oxidation number of -2, we can calculate the oxidation number of manganese (Mn). Therefore, (+7) + x + (-8) = -1, solving for x we get +6.
SO3-2: The overall charge of the sulfite ion is -2. Since oxygen is typically assigned an oxidation number of -2, sulfur (S) must have an oxidation number of +4.
SO4-2: The overall charge of the sulfate ion is -2. Since oxygen is typically assigned an oxidation number of -2, sulfur (S) must have an oxidation number of +6.
Mn+2: Since Mn forms a cation with a +2 charge, its oxidation number is +2.
Step 2: Divide the equation into two half-reactions: one for the oxidation and one for the reduction. In this equation, manganese (Mn) is undergoing reduction, while sulfur (S) is undergoing oxidation.
Half-reaction for the reduction: MnO4- + 8H+ + 5e- ===> Mn+2 + 4H2O
Half-reaction for the oxidation: SO3-2 ===> SO4-2
Step 3: Balance the atoms other than hydrogen and oxygen in each half-reaction.
For the reduction half-reaction:
MnO4- + 8H+ + 5e- ===> Mn+2 + 4H2O
Balance Mn atoms: Add 1 Mn to the right side.
MnO4- + 8H+ + 5e- ===> Mn+2 + 4H2O
For the oxidation half-reaction:
SO3-2 ===> SO4-2
Balance S atoms: Add 1 S to the right side.
SO3-2 ===> SO4-2
Step 4: Balance the charges in each half-reaction by adding electrons (e-).
For the reduction half-reaction:
MnO4- + 8H+ + 5e- ===> Mn+2 + 4H2O
For the oxidation half-reaction:
SO3-2 + 2e- ===> SO4-2
Step 5: Multiply each half-reaction by the necessary coefficients to equalize the number of electrons transferred.
To balance the electrons, multiply the reduction half-reaction by 2:
2(MnO4- + 8H+ + 5e-) ===> 2(Mn+2 + 4H2O)
The final balanced equation is:
2MnO4- + 16H+ + 10e- ===> 2Mn+2 + 8H2O
3SO3-2 + 6e- ===> 3SO4-2
Step 6: Add the two balanced half-reactions together to obtain the balanced overall equation.
2MnO4- + 16H+ + 10SO3-2 ===> 2Mn+2 + 8H2O + 10SO4-2
And there you have it! The equation MnO4- + SO3-2 ===> SO4-2 + Mn+2 is balanced.