Balance Equations Using Half-Cell Method

[ipl]

Balance the following equations by the half-cell method.Show both half-cell reactions and identify them as oxidation or reduction.
a) SO_3^2- + MnO_4^- +H^+ <- ->Mn^2+ SO_4^2- +H_2O(l)
b) Cl_2(g) + OH^- <- -> Cl^- + ClO_3^- +H_2O(l)
c) SO_4^2- +I^- +H+ <- -> S^2- + I_2(s) + H_2O (l)

I honestly don't know where to start so if you could show me all the steps, I would greatly appreciate it. Thanks in advance.

 

[ikjadoon]

Hey, I just did this in class like two days ago! :)

All right, I need a little background. How much do you know? Oxidation vs reduction? OILRIG? Anything?

The quick way:

1. Find out what's oxidized and what's reduced: calculate the oxidation # for each atom in the reactant side and then compare that with the oxidation # for each atom on the product side. If something has a loss of electrons, it's oxidized. If it something has a gain of electrons, it's reduced. Now, there are certain "rules" about oxidation that should be listed in your chemistry book (if you have one). One of them is that oxygen always has a charge of (2-) except in peroxides (O-O single bond). I mention these because this way, you don't HAVE to compare the oxidation # of every single atom on both sides, just the ones that are likely to change.
2. Write an equation that shows the JUST oxidation/reduction of each reactant. So for example:
   
   SO_3^2- <<<<<<------>>>>>> SO_4^2-
   
   Now remember, you need one for the sulfate and also one for the permanganate. But you see how the mass on the equation above is NOT balanced? Balance that by adding water (H_2O) to the oxygen-deficient side. For every water added on the oxygen-deficient side, add two H+ ions on the other side. This is "allowed" because these reactions are usually done in water, so we can grab H+s and H_2Os from the solution. Doing this will also keep the mass balanced. :)
   
   Now you'll get an equation with the MASSES balanced, but the charges not. Looking at the sulfur half-cell reaction: include electrons on whichever side it is needed (on the positive side to make the equation balance charge-wise in terms of the sulfur). You might think "the charge is equal on both sides!" Yes, the OVERALL charge is the same on both sides, but look at the charge of the sulfur. See how it's changed, if O is always (2-)? Do this exact same thing for the permanganate. You will now have two equations which are your balanced half-cell equations, balanced in both charge and mass.
   
   Add the two equations together. If the electron counts do NOT cancel, make them. :) By that I mean multiplying one (or maybe both) reactions by some coefficients.

 

[Blueshift5]

The half-cell method is a systematic way to balance redox equations by breaking them down into two half-reactions: the oxidation half-reaction and the reduction half-reaction. These half-reactions are then balanced individually and combined to give the overall balanced equation.

a) SO3^2- + MnO4^- +H+ -> Mn^2+ + SO4^2- +H2O(l)

Oxidation half-reaction: SO3^2- -> SO4^2- + 2e-
Reduction half-reaction: MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O

Combining the two half-reactions, we get:

SO3^2- + MnO4^- + 8H+ -> SO4^2- + Mn^2+ + 4H2O

b) Cl2(g) + OH- -> Cl- + ClO3^- +H2O(l)

Oxidation half-reaction: 2Cl- -> Cl2 + 2e-
Reduction half-reaction: Cl2 + 2H2O + 2e- -> 2ClO3^- + 4H+

Combining the two half-reactions, we get:

2Cl2 + 2H2O + 4OH- -> 4Cl- + 4ClO3^- + 4H2O

c) SO4^2- + I^- +H+ -> S^2- + I2(s) + H2O (l)

Oxidation half-reaction: I^- -> I2 + 2e-
Reduction half-reaction: SO4^2- + 4H+ + 2e- -> S^2- + 2H2O

Combining the two half-reactions, we get:

SO4^2- + I^- + 4H+ -> S^2- + I2 + 2H2O

In all three equations, the species that loses electrons is undergoing oxidation, and the species that gains electrons is undergoing reduction. It is important to note that the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction in order for the equation to be balanced. Practice makes perfect, so keep practicing balancing equations using the half-cell method