BALANCE USING HALF-REACTIONS
ACIDIC CONDITIONS:
S(s) + NO3-(aq)  SO2 (g) + NO(g)
Step 1: Start by breaking them into half reactions
S(s)

NO3-(aq)
SO2 (g)

NO(g)
For ease, I will balance the half-reactions one at a time, starting with
sulfur.
S(s)

SO2 (g)
Step 2: Balance the atoms (S). This is already done.
Step 3: Balance for oxygen using H2O(l)
2 H2O(l)
+
S(s)

SO2 (g)
Step 4: Balance for hydrogen using H+
2 H2O(l)
+
S(s)

SO2 (g) + 4 H+
Step 5: Balance for charge using electrons.
2 H2O(l)
+
S(s)

SO2 (g) + 4 H+
left = 0
right = + 4
So, we must add four electrons to the right side.
2 H2O(l)
+
S(s)

SO2 (g) + 4 H+ + 4 e-
Step 6: Balance the other half reaction and add them together.
NO3-(aq)
4 H+ + NO3-(aq)

NO(g) + 2 H2O(l)

NO(g) + 2 H2O(l)
3 e- + 4 H+ + NO3-(aq) 
NO(g) + 2 H2O(l)
ADD:
2 H2O(l)
+
S(s)

3 e- + 4 H+ + NO3-(aq) 
SO2 (g) + 4 H+ + 4 eNO(g) + 2 H2O(l)
If we tried to add the reactions right now, we would see that the
electrons do not cancel out, so we must multiply the sulfur reaction by
3 and the nitrogen reaction by 4.
3
*
(2 H2O(l)
+
S(s)
4
*
(3 e- + 4 H+ + NO3-(aq) 

SO2 (g) + 4 H+ + 4 e- )
NO(g) + 2 H2O(l) )
6 H2O(l)
+ 3
S(s)
 3 SO2 (g) + 12 H+ + 12 e-
12 e- + 16 H+ + 4 NO3-(aq)
 4 NO(g) + 8 H2O(l)
6 H2O(l) + 3 S(s) + 16 H+ + 4 NO3-(aq)  3 SO2 (g) + 12 H+ + 4 NO(g) + 8 H2O(l)
Now we can cancel out terms that appear on both sides.
3 S(s) + 4 H+ + 4 NO3-(aq)  3 SO2 (g) + 4 NO(g) + 2 H2O(l)
CHECK: MAKE SURE YOU CHECK THAT THE CHARGES AND
ATOMS BALANCE!!!
BASIC CONDITIONS
Cr(OH)3 (s) + ClO3- (aq)  CrO42-(aq) + Cl-(aq)
Step 1: Break into half reactions.
Cr(OH)3 (s)  CrO42-(aq)
ClO3- (aq)  + Cl-(aq)
As before, I will start with the chromium half reaction and balance it
and then move onto the chlorine.
Step 2: Balance for atoms, which is done.
Step 3: Balance for oxygen using H2O(l)
H2O(l) + Cr(OH)3 (s)  CrO42-(aq)
Step 4: Balance for hydrogen using H+
H2O(l) + Cr(OH)3 (s)  CrO42-(aq) + 5 H+
Step 5: Add one OH- to BOTH sides for every H+ ion present.
5 OH- + H2O(l) + Cr(OH)3 (s)  CrO42-(aq) + 5 H+ + 5 OHCombine the OH- and H+ to form water.
5 OH- + H2O(l) + Cr(OH)3 (s)  CrO42-(aq) + 5 H2O(l)
Then cancel out the water on the other side.
5 OH- + Cr(OH)3 (s)  CrO42-(aq) + 4 H2O(l)
Step 6: Balance for charge using electrons.
5 OH- + Cr(OH)3 (s)  CrO42-(aq) + 4 H2O(l)
left = - 5
right = - 2
So, we should add three electrons to the right side.
5 OH- + Cr(OH)3 (s)  CrO42-(aq) + 4 H2O(l) + 3 e-
Step 7: Balance the other half-reaction and add together.
ClO3- (aq)  + Cl-(aq)
ClO3- (aq)  + Cl-(aq) + 3 H2O(l)
6 H+ + ClO3- (aq)  + Cl-(aq) + 3 H2O(l)
6 OH- + 6 H+ + ClO3- (aq)  + Cl-(aq) + 3 H2O(l) + 6 OH6 H2O(l) + ClO3- (aq)  + Cl-(aq) + 3 H2O(l) + 6 OH3 H2O(l) + ClO3- (aq)  + Cl-(aq) + 6 OH6 e- + 3 H2O(l) + ClO3- (aq)  + Cl-(aq) + 6 OH-
ADD:
5 OH- + Cr(OH)3 (s)  CrO42-(aq) + 4 H2O(l) + 3 e6 e- + 3 H2O(l) + ClO3- (aq)  + Cl-(aq) + 6 OHBalance the electrons
10 OH- + 2 Cr(OH)3 (s)  2 CrO42-(aq) + 8 H2O(l) + 6 e6 e- + 3 H2O(l) + ClO3- (aq)  + Cl-(aq) + 6 OH-
10 OH- + 2 Cr(OH)3 (s) + 3 H2O(l) + ClO3- (aq)  2 CrO42-(aq) + 8 H2O(l) + Cl-(aq) + 6 OH-
Now we can cancel out terms that appear on both sides.
4 OH- + 2 Cr(OH)3 (s) + ClO3- (aq)  2 CrO42-(aq) + 5 H2O(l) + Cl-(aq)
CHECK: MAKE SURE YOU CHECK THAT THE CHARGES AND
ATOMS BALANCE!!!