Write overall reaction step 3 remember these occur in

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write overall reactionStep 3? – remember these occur in water; balance oxygen atoms by addingH2o and hydrogens by adding H+-2+64H2O + CuS Cu2+ SO4 2- + 8H+ (because it occurs in water) 8e- (must add 8 electrons so it’s balanced) NO3-(aq) + 4H+ + 3e- NO(g) + 2H2O
Multiply top one by 3, latter by 8, add together then add together 1/29/16Balance Redox Reations (Half Reaction MethodoCuS(s) + NO3-(aq) Cu2+(aq) + SO4 2-(aq) + NO(g)Oxygen remains unchanged (-2), whereas S and Nitrogen undergo changesCopper remains the same (+2)oWrite out the half reactions (not balanced, but with key compounds or ions present)Oxidation: (balance out atoms and chargesCuS(s) Cu2+(aq) + SO4 2-(aq) oxidation (S changes from -2 to +6 = oxidation)NO3-(aq) NO(g) reduction (N goes from +5 to +2)oStep 3: balance oxygen atoms by adding H2O and hydrogens by adding H+ CuS(s) + 4H2O(l)Cu2+(aq) + SO4 2-(aq) + 8H+(aq)NO3-(aq) + 4H+(aq) NO(g) + 2H2O(l)oStep 4: balance net charges by adding electronsOxidation: CuS(s) + 4H2O(l)Cu2+(aq) + SO4 2-(aq) + 8H+(aq) + 8e-Left side is completely neutralNO3-(aq) + 4H+(aq) NO(g) + 2H2O(l)Would add 2 waters on the right to even out the oxygen moleculesSo you’d have to add 4H+ on the left side; atoms are all balanced, add electronsReduction: NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)Once you combine these two you should get the overall reactionoStep 5 The number ofe electrons should match up, so multiply both sides by a common factor.. multiply (ox) by 3: 12H2O + 3CuS 3Cu2+ + 3SO4 2- + 24e- + 24H+multiply (red) by 8: 8NO3- + 32H+ + 24e- 8NO + 16H2Oostep 6: add these two together and cancel stuff to get overall balanced equation: 8NO3- + 3CuS + 8H+ 8NO + 3SO4 2- + 3Cu2+ + 4H2O = overall balanced redox reaction in acidic media oadd required number of hydroxide ions to both side such that you neutralize all the acid present; generating equal amounts of hydroxide ions (ex. If you add 8 OH- to left, then add 8OH to left), by cancelling out 8 water molecuels with 4 water molecuels… acidc/basic media) oStep 7: solve for acidc media frsit For basic media (solutions)Add OH- to both sides for every H+, H+ + OH- becomes H2O, cancel water molecules3CuS + 8NO3- + 8H2O 3CuPracticeoAcid solution: VO2++ Zn VO2++ Zn2+
Zn getting oxidized (goes from 0 to 2+)Oxygen stays the same, whereas V goes from (+5 to +4) gets reducedStep 1-3: Oxidation: Zn Zn2++ 2e-Reduction: VO2++ 2H+ + e- VO2++ H2OMultiply it: Oxidation: Zn Zn2+ + 2e-Reduction: 2VO2+ + 4H+ + 2e- 2VO2+ + 2H2OAdd, and you get: 2VO2++ Zn + 4H+2VO2++ 2H2O + Zn2+ oBalance the following reaction in basic mediaCl2(g) Cl-(aq) + OCl-(aq)Cl goes from 0 to -1 in Cl- and from 0 to +1 in OCl-Oxidation: Cl2 +2H2O 2OCl- + 4H+ + 2e-Reduction: Cl2 +2e- 2Cl-Add together: 2H2O + 2Cl2 2Cl- + 2OCl- + 4H+ Solved in acidic media, you need to add required OH- to neutralizeH+2H2O + 2Cl2 + 4OH- 2Cl- + 2OCl- + 4H+ + 4OH-2H2O + 2Cl2 + 4OH- 2Cl- + 2OCl- + 4H2Ofinal answer: 4OH + 2Cl2 2Cl- + 2OCl- + 2H2O (divide everything by 2) Electrochemistry: electrochemical cells and cell partsRedox reactionsoClassifications: Direct – no external circuitIndirect – external circuit o

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