Question

Part IV: Balancing Redox Reactions in Acidic and Basic Solution General rules for balancing redox reactions in acidic or basic solution: 1. Divide the redox reaction into two half-reactions. One that contains the element that gets Oxidized and one that contains the element that gets reduced. 2. Focusing on one half-reaction at a time, balance all non-H and non-0 atoms. 3. Balance the O atoms by adding water molecules to the side with too few oxygens 4. Balance the H atoms by adding Hions to the side with too few hydrogens. 5. Balance charge by adding electrons (e-) to the side with the higher charge. (This should be the left for the reduction and the right for the oxidation.) 6. Repeat steps 2-5 for the other half-reaction. 7. Combine the two half reactions by multiplying one or both half-reactions by an integer so that the electrons will cancel when the two reactions are added together. 8. Check that the atoms and charges are balanced. 9. If the reaction is in acid solution, then the reaction is balanced. If the reaction is in basic solution, then add one OH-to BOTH sides of the reaction for every Hi present. 10. Combine the H+ ions and OH-ions to form water. Simplify by canceling any excess waters from both sides of the reaction. 11. Check again that the atoms and charges are balanced. Balance each of the following redox reactions in both acidic and basic solution using the half reaction method outlined above. 10. Unbalanced: HNO2(aq) + Al(s) + NH3(0) + HAIO:(aq) 11. Unbalanced: C032 (aq) + NO(g) - CO(g) + NO3- (aq) Recitation 8

Answer 1

I change ofc in (032- (11) 03 (6) + No copt login M-6= -2 n=t4 +5 +2 (140 27+ No-2 (90) +2 cor+ No (92 aw change of N in Noz- KE -176 1511 oxidation No I Noz Noz t3e- Reduction coza- co I co, 2 + 2e - - No Co n NO + 2H₂0 - Noz + 3e - coz + 20 Co NO + 2H₂0 - Noz + 4H+ + 36 - co +20" Co + 2 H2O 2 Calo + 2 H₂O Not tunt +36-) $ 42 Not 4160 2002 +8 H+ +66 - Coz-te +4H* - Cot 2H2O Mere both oxidation and reduction |3 (103 + 2€ +4+* CO + 2H20] half reaction multiplied by 2 and 3 to equalise changel 300,31 +62 +12 #- 360 +6140 Adding both ost Will 2 NO +3603" +4H+ 2 No." + 300+ 21,0 in Basic Solution, 2 NO + 3603 + 4 OH" +447 - 82N103* +300 + 211,0 + 4014 – + 2 N03 + 300 + 2 H20 + 4014 - 2 No + 3603 + H2O H2o get Cancelled +2 HO .: 200 + 300 & 2NO3 + 300 + 40H- Not 3008 + 4H20 2003 + 400 + 40H -

10) Hala aus + Also le + HALO2 (8) HNO2 1900 + Als, Milly + 14 Ale 02 The change found by Considering a Species is neutral. The elemental stale Al has 2e no change molecular oxidation Reduction HAO2 - Allz Als, HALO Als, HALO₂ + 3e - HNO, + 6e" - NH3 14116,+ be" -- 11 Hz + 2 H2O 3 Als, +240 - HAIG+ + + 36- BORA multiply whole equation by 2 to balance electron Hinheat Gata HNot 6e +6H7 ~ BNH ₂ + 2 H₂O 2 Al + 4H₂O → 2 Halo +84 the" Adding oxidation Reduction half cell 2A1 + HNO, + 2H20 - 2HA 102 + NIH, er lence Balanced equation 2 Al + Hne, + 211,0 - 2 Halot Nity