Step 6 Balance H’s, by adding 4 H+ to right side of 2nd equation.
2 H2O + SO2 → SO4 -2 + 4 H+
Step 7 Balance charges by adding 2 e- to left side of 1st equation and 2 e- to right side of 2nd equation
2e- + Br2 → 2Br -1
2 H2O + SO2 → SO4 -2 + 4 H+ + 2 e-
Step 8, electrons are balanced
Step 9, add the 2 half-reactions and subtract any duplications.
Br2 + 2 H2O + SO2 → 2 Br- + SO4^2- + 4 H+
2 Br = 2 Br, 4 H = 4 H, 2 O + 2O = 4 O
-2 + -2 + +4 = 0
The equation is balanced.
If the reaction occurs in a basic solution, add the same number of OH-1ions to both sides to equalize the number of H+1 ions.
So add 4 OH-1 to both sides
4 OH-1 + Br2 + 2 H2O + SO2 → 2 Br- + SO4^2- + 4 H+ + 4 OH-1
4 H+ + 4 OH-1 = 4 H2O
4 OH-1 + Br2 + 2 H2O + SO2 → 2 Br- + SO4^2- + 4 H2O
Subtract 2 H2O from both sides
4 OH-1 + Br2 + SO2 → 2 Br- + SO4^2- + 2 H2O
4 O + 2 O = 4 O + 2 O
2 Br = 2 Br
1 S = 1 S
-4 charge = 2 -1 charges + -2 charge
The equation is balanced in a basic solution!
This method works well for oxidation reduction reactions!
You might wonder what positive ion would be in the original basic solution!
Sodium hydroxide is the most common base.
Let’s add 4 Na+1 ions to both sides!
4 Na+1 + 4 OH-1 + Br2 + SO2 → 4 Na+1 + 2 Br- + SO4^2- + 2 H2O
4 NaOH (aq) + Br2 + SO2 → 2 NaBr (aq) + Na2SO4 (aq) + 2 H2O
(aq) means the compound is dissolved in water.
NaOH, NaBr, and Na2SO4 will ionize when dissolved in water. So, you might say that the 4 Na+1 ions were “spectators”, since Na did not react during the reaction!