Balancing Redox reactions in an acid or a

Balancing Redox reactions in an acid or a base

Redox reactions in acidic solutions o o o o I will tell you if it is in an acidic solution. These have special rules. Separate the reaction into half reactions. Balance all elements except hydrogen and oxygen. Balance oxygen by adding H 2 O (which is always prevalent in an acidic solution) Balance hydrogen by adding H+. Then balance the charge by adding electrons to whichever side is more positive. Recombine your two half equations.

Example o o o In an acidic solution Cr 2 O 7 2 - + Cl- → Cr 3+ + Cl 2 Half reactions Cr 2 O 7 2 - → Cr 3+ Cl- → Cl 2

Here we go o o Cr 2 O 7 2 - → Cr 3+ Cr 2 O 7 2 - → 2 Cr 3+ + 7 H 2 O Cr 2 O 7 2 - + 14 H+→ 2 Cr 3+ + 7 H 2 O Cr 2 O 7 2 - + 14 H++ 6 e- → 2 Cr 3+ + 7 H 2 O

Other side o o o Cl- → Cl 2 2 Cl- → Cl 2 + 2 e. I have to equal 6 e- so multiply by 3 6 Cl- → 3 Cl 2 + 6 e-

Combine my half reactions o o o Cr 2 O 7 2 - + 14 H++ 6 e- → 2 Cr 3+ + 7 H 2 O 6 Cl- → 3 Cl 2 + 6 e. And you get Cr 2 O 7 2 - +14 H++ 6 Cl- → 2 Cr 3+ + 3 Cl 2 + 7 H 2 O The electrons cancel out.

Balance in an acidic solution o NO 2 + Cl. O 3 - NO 3 - + Cl 2

Balance in an acidic solution o o NO 2 + Cl. O 3 - NO 3 - + Cl 2 Half reactions NO 2 NO 3 Cl. O 3 - Cl 2

Nitrate side o o NO 2 NO 3 NO 2 + H 2 O 2 H+ + NO 3 - + 1 e-

Chlorate o Cl. O 3 - Cl 2 2 Cl. O 3 - Cl 2 + 6 H 2 O 2 Cl. O 3 - + 12 H+ + 10 e- Cl 2 + 6 H 2 O o You will have 10 x the first reaction o 10 NO 2 +10 H 2 O 20 H+ +10 NO 3 - + 10 e- o o

Put them together o o 2 Cl. O 3 - + 12 H+ + 10 NO 2 +10 H 2 O Cl 2 + 6 H 2 O + 20 H+ +10 NO 3 Notice the H+ and the water can also cancel out 2 Cl. O 3 - + 10 NO 2 +4 H 2 O Cl 2 + 8 H+ +10 NO 3 -

Example o o In an acidic solution Mn. O 4 - + H 2 O 2 → Mn 2+ + O 2

Example o o o In an acidic solution Mn. O 4 - + H 2 O 2 → Mn 2+ + O 2 Half reactions Mn. O 4 - → Mn 2+ H 2 O 2 → O 2

Top Equation o o Mn. O 4 - → Mn 2+ + 4 H 2 O Mn. O 4 - + 8 H+→ Mn 2+ + 4 H 2 O Mn. O 4 - + 8 H++ 5 e-→ Mn 2+ + 4 H 2 O

Bottom Equation o o o o H 2 O 2 → O 2 + 2 H+ + 2 e. I need to equal 5 e- so… That won’t work… 2 Mn. O 4 - + 16 H++ 10 e-→ 2 Mn 2+ + 8 H 2 O 5 H 2 O 2 → 5 O 2 + 10 H+ + 10 e-

Add them together o o o 2 Mn. O 4 - + 16 H++ 10 e-→ 2 Mn 2+ + 8 H 2 O 5 H 2 O 2 → 5 O 2 + 10 H+ + 10 e. And you get 2 Mn. O 4 - + 6 H++ 5 H 2 O 2 → 2 Mn 2+ + 5 O 2 + 8 H 2 O Notice the H+ canceled out as well.

Balancing Redox Equations in a basic solution o o Follow all rules for an acidic solution. After you have completed the acidic reaction add OH- to each side to neutralize any H+. Combine OH- and H+ to make H 2 O. Cancel out any extra waters from both sides of the equation.

Example o o o We will use the same equation as before In a basic solution Mn. O 4 - + H 2 O 2 → Mn 2+ + O 2 Balanced in an acidic solution 2 Mn. O 4 - + 6 H++ 5 H 2 O 2 → 2 Mn 2+ + 5 O 2 + 8 H 2 O

Basic solution o o Since this is a basic solution we can’t have excess H+. We will add OH- to each side to neutralize all H+ 2 Mn. O 4 - + 6 H++ 5 H 2 O 2 + 6 OH→ 2 Mn 2+ + 5 O 2 + 8 H 2 O + 6 OHWe added 6 OH- because there were 6 H+

Cont. o o o H+ + OH- → H 2 O Combine the hydroxide and hydrogen on the reactant side to make water 2 Mn. O 4 - + 6 H 2 O + 5 H 2 O 2 → 2 Mn 2+ + 5 O 2 + 8 H 2 O + 6 OHCancel out waters on both sides 2 Mn. O 4 - + 5 H 2 O 2 → 2 Mn 2+ + 5 O 2 + 2 H 2 O + 6 OH-

Another example o o In a basic solution Mn. O 4 − + SO 32 -→Mn. O 4 2− + SO 42 -

Another example o o o In a basic solution Mn. O 4 − + SO 32 -→Mn. O 4 2− + SO 42 Half reactions Mn. O 4 − → Mn. O 4 2− SO 32 -→ SO 42 -

Half reactions o o o o Mn. O 4 − → Mn. O 4 2− Mn. O 4 - + e- → Mn. O 4 2− SO 32 -→ SO 42 H 2 O + SO 32 -→ SO 42 - + 2 H+ +2 e. Double the top reaction

o o o 2 Mn. O 4 - + 2 e- → 2 Mn. O 4 2− H 2 O + SO 32 -→ SO 42 - + 2 H+ +2 e. Combine them 2 Mn. O 4 - + H 2 O + SO 32→ 2 Mn. O 4 2− +SO 42 - + 2 H+ Add OH 2 Mn. O 4 - + H 2 O + SO 32 - + 2 OH→ 2 Mn. O 4 2− +SO 42 - + 2 H+ + 2 OH-

o o o 2 Mn. O 4 - + H 2 O + SO 32 - + 2 OH→ 2 Mn. O 4 2− +SO 42 - + 2 H 2 O finishing 2 Mn. O 4 - + SO 32 - + 2 OH→ 2 Mn. O 4 2− +SO 42 - + H 2 O