Chapter 5 Oxidation–Reduction Reactions Chemistry: The Molecular Nature of Matter, 7E Jespersen/Hyslop

Oxidation Reduction Reaction Oxidizing Agent (Oxidizer) Substance that accepts electrons – Accepts electrons from another substance – Substance that is reduced – Cl e –  2Cl – Reducing Agent (Reducer) Substance that donates electrons – Releases electrons to another substance – Substance that is oxidized – Na  Na + + e – 2

REDOX in Aqueous Solution Example: Mix solutions of K 2 Cr 2 O 7 and FeSO 4 – Dichromate ion, Cr 2 O 7 2–, oxidizes Fe 2+ to Fe 3+ – Cr 2 O 7 2– is reduced to form Cr 3+ – Acidity of mixture decreases as H + reacts with oxygen to form water Skeletal Eqn. Cr 2 O 7 2– + Fe 2+  Cr 3+ + Fe 3+ 3

Ion Electron Method Balance in Acidic Solution (H + ions Present) Cr 2 O 7 2– + Fe 2+  Cr 3+ + Fe Break into half-reactions Cr 2 O 7 2–  Cr 3+ Fe 2+  Fe Balance atoms other than H and O Cr 2 O 7 2–  2Cr 3+ Put in 2 coefficient to balance Cr Fe 2+  Fe 3+ Fe already balanced 4

Ion-Electron Method in Acid 3. Balance O by adding H 2 O to the side that needs O Cr 2 O 7 2–  2Cr 3+ Left side has seven O atoms Right side has none Add seven H 2 O to right side Fe 2+  Fe 3+ No O to balance 5 + 7H 2 O

Ion-Electron Method in Acid 4. Balance H by adding H + to side that needs H Cr 2 O 7 2–  2Cr H 2 O Right side has fourteen H atoms Left side has none Add fourteen H + to left side Fe 2+  Fe 3+ No H to balance 6 14H + +

Ion-Electron Method in Acid 5. Balance net charge by adding electrons. 14H + + Cr 2 O 7 2–  2Cr H 2 O 6 electrons must be added to reactant side Fe 2+  Fe 3+ 1 electron must be added to product side Now both half-reactions balanced for mass and charge 7 6 e – + + e – Net Charge = 2(+3)+7(0) = 6 Net Charge = 14(+1) +(–2) = 12

6. Make electron gain equal electron loss; then add half-reactions 6 e – + 14H + + Cr 2 O 7 2–  2Cr H 2 O Fe 2+  Fe 3+ + e – 7. Cancel anything that's the same on both sides Ion-Electron Method in Acid 6[6[ ] 8 6Fe Cr H 2 O + 6 e – 6 e – + 6 Fe H + + Cr 2 O 7 2–  6Fe H + + Cr 2 O 7 2–  6Fe Cr H 2 O

Ion-Electron in Basic Solution The simplest way to balance an equation in basic solution Use steps 1 – 7 above, then 8. Add the same number of OH – to both sides of the equation as there are H + 9. Combine H + and OH – to form H 2 O 10. Cancel any H 2 O that you can from both sides 9

Ion-Electron in Basic Solution Returning to our example of Cr 2 O 7 2– and Fe Add to both sides of equation the same number of OH – as there are H +. 9.Combine H + and OH – to form H 2 O. 10. Cancel any H 2 O that you can 10 6Fe H + + Cr 2 O 7 2–  6Fe Cr H 2 O + 14 OH – 6Fe H 2 O + Cr 2 O 7 2–  6Fe Cr H 2 O + 14OH – 7 6Fe H 2 O + Cr 2 O 7 2–  6Fe Cr OH –

Your Turn! Which of the following is a correctly balanced reduction half-reaction? A. Fe 3+ + e –  Fe B. 2Fe + 6HNO 3  2Fe(NO 3 ) 3 + 3H 2 C. Mn H 2 O  MnO 4 – + 8H e – D. 2O 2–  O e – E. Mg e –  Mg 11

Your Turn! Balance the following reaction in acidic solution I – + HNO 2  NO + I 2 A.H + + HNO 2 + I –  NO + I 2 + H 2 O B.2H + + 2HNO 2 + I –  2NO + I 2 + 2H 2 O C.2H + + 2HNO 2 + 2I –  2NO + I 2 + 2H 2 O D.H + + HNO 2 + 2I –  2NO + I 2 + 2H 2 O E.H + + HNO 2 + 2I –  NO + I 2 + H 2 O + e – 12

Ion-Electron in Basic Solution Balance the following equation in basic solution: MnO 4 – + HSO 3 –  Mn 2+ + SO 4 2– 1. Break it into half-reactions MnO 4 –  Mn 2+ HSO 3 –  SO 4 2– 2. Balance atoms other than H and O MnO 4 –  Mn 2+ Balanced for Mn HSO 3 –  SO 4 2– Balanced for S 13

Ion-Electron in Basic Solution 3.Add H 2 O to balance O MnO 4 –  Mn 2+ HSO 3 –  SO 4 2– 4.Add H + to balance H MnO 4 –  Mn H 2 O H 2 O + HSO 3 –  SO 4 2– H 2 O H 2 O + 8H H +

Ion-Electron in Basic Solution 5.Balance net charge by adding electrons. 8H + + MnO 4 –  Mn H 2 O 8  (+1) + (–1) = = +2 Add 5 e – to reactant side H 2 O + HSO 3 –  SO 4 2– + 3H (–1) = –1 –2 + 3(+1) = +1 Add 2 electrons to product side 5e– +5e– e – 15

Ion-Electron in Basic Solution 6.Make electron gain equal electron loss 5 e – + 8H + + MnO 4 –  Mn H 2 O H 2 O + HSO 3 –  SO 4 2  + 3H e – – Must multiply Mn half-reaction by 2 – Must multiply S half-reaction by 5 – Now have 10 electrons on each side 16 2[2[ ] 5[5[ ]

6.Then add the two half-reactions 10e – + 16H + + 2MnO 4   2Mn H 2 O 5H 2 O + 5HSO 3   5SO 4 2  + 15H e – 7. Cancel anything that is the same on both sides. Balanced in acid H + + 2MnO 4 + 5HSO 3 – 2Mn H 2 O + 5SO 4 2–  10 e – + 16H + + 2MnO 4  + 5H 2 O + 5HSO 3  2Mn H 2 O + 5SO 4 2  + 15H e –  Ion-Electron in Basic Solution 17 31

Ion-Electron in Basic Solution 8. Add same number of OH – to both sides of equation as there are H + 9.Combine H + and OH – to form H 2 O 10. Cancel any H 2 O that you can 2MnO 4 – + 5HSO 3 –  2Mn H 2 O + OH – + 5SO 4 2– + OH – 18 H + + 2MnO 4 – + 5HSO 3 – 2Mn H 2 O + 5SO 4 2–  + OH – H 2 O + 2MnO 4 – + 5HSO 3 – 2Mn H 2 O + 5SO 4 2  + OH –  2

19 Balance each equation in the solution indicated by the skeletal reaction using the ion electron method. MnO 4 – + C 2 O 4 2–  MnO 2 + CO 3 2– + H + in acid 2MnO 4 – + 3C 2 O 4 2– + 2H 2 O  2MnO 2 + 4H + + 6CO 3 2– ClO – + VO 3 –  ClO 3 – + V(OH) 3 + OH – in base ClO – + 4H 2 O + 2VO 3 –  ClO 3 – + 2V(OH) 3 + 2OH – Your Turn!