Balancing redox reactions

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Balancing redox reactions

  1. 1. SUALEHA IQBAL
  2. 2. REDOX REACTIONS: “ Transfer of electrons between two species.” INVOLVE TWO TYPES OF AGENTS: OXIDING AGENT: A compound that reduced is refer to as oxidizing agent. EXAMPLES: Br2 ,I2 , H2O2, KMnO4. REDUCING AGENT: A compound that is oxidize is refer to as reducing agent. EXAMPLES: I- ,H2S, Zn.2
  3. 3. Types Of Species Involve In Redox Reactions: OXIDIZE SPECIE REDUCE SPECIE3
  4. 4. 4
  5. 5. IDENTIFICATION OF REDOX REACTIONS: If the reaction is redox than it involve transfer of electrons from one atom, ion or molecule to another i.e. change in oxidation state (can observe through change in color).5
  6. 6. OXIDATION STATE “It is the way to describe the number of electrons that have been transferred or shared between atoms of different kind.” Hydrogen =1 (except hydrides). Oxygen=(-2). Metal (+). Compound = 0.If two elements in a reaction change in oxidation state, one increasing,the other decreasing,Then the reaction is redox
  7. 7. BALANCING REDOX REACTIONS: From ion electron method: For example MnO4- (aq) + I- (aq) Mn2+ (aq) + I2(s) •Identification of oxidize and reduce species: Manganese (Mn) goes from a charge of +7 to a charge of +2. Since it gained 5 electrons, it is being reduced. Next we see that Iodine (I) goes from a charge of -1 to 0. Thus it lost electrons, and is being oxidized. •Divide equation in half reactions: First half is the reduction half: MnO4- (aq) Mn2+ (aq) Next we have the oxidation half: I-(aq) I2 (s)7
  8. 8. BALANCING PART: •Four water molecules for 4 oxygen in MnO4. •Balance hydrogen by adding hydrogen to other side Reduction half: MnO4- (aq) + 8H+ + 5e- Mn2+ (aq) +4H2O •5 electrons on the side having +7 charge are added to balance charge. Reduction half is balance. • Now come to oxidation half equation: •In The oxidation half is iodine, so we can balance it easily by adding another iodine to the left side. OXIDATION HALF : 2I- (aq) I 2(S)+2(e-)8
  9. 9. • Multiply reduction half cell by 2 and oxidation half equation by 5 for balancing complete equation. 2{MnO4-(aq)+8H +5e- Mn+2(aq) +H2O} 5{2I- (aq) I2(s) + 2e-} Reduction half becomes: 2MnO4- (aq) + 16H+ (aq) + 10e- 2Mn2+ (aq) + 8H2O (l) Oxidation half becomes: 10I- (aq) 5I2 (s) + 10e-.9
  10. 10. Adding both equations : 2MnO4- (aq) + 16H+ (aq) + 10e- 2Mn2+ (aq) + 8H2O (l) 10I- (aq) 5I2 (s) + 10e- 10I- (aq) + 2MnO4- (aq) + 16H+ (aq) 5I2 (s) + 2Mn2+ (aq) + 8H2O (l) Final inspection : Proton : Oxygen: Charge: Balanced10
  11. 11. 2nd Example Through oxidation state method: EQUATION : Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O •Redox or not: (write oxidation states under equation) Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O 0 +5 +2 +2 • Insert co-efficient so that the total decrease in oxidation state of one Element equals the total increase in oxidation state of other element. +2 ×3=+6 Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O -3×2= -6 3Cu (S)+2HNO3(aq 3Cu+2(aq)+ 2NO (g)11
  12. 12. •Balance oxygen: 4H2O on right will balance oxygen. •Balance hydrogen by adding 6protons on left side. 3Cu (S)+2HNO3(aq)+6H+ 3Cu+2(aq)+ 2NO (g) +4H2O •Spectators ions: 6 NO3- on each side of equation 3Cu (S)+8HNO3(aq) 3Cu(NO3)2 (aq)+ 2NO (g) +4H2O(l) BALANCED12
  13. 13. Balancing organic redox reactions: oxidation of n-butyl alcohol with K2Cr2O7 CH3CH2CH2CH2-OH + K2Cr2O7 , H+ CH3CH2CH2COOH half-reactions for the oxidation and the reduction involved: oxidation: CH3CH2CH2CH2-OH CH3CH2CH2COOH reduction: Cr2O72- Cr3+ Mass balance: oxidation: CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4 H+ reduction: Cr2O72- + 14H+ 2 Cr3+ + 7 H2O13
  14. 14. Charge balance oxidation: CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4H+ + 4e- reduction: 6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O No. of electron in oxidation half = No. of electron in reduction half • 3 x{( CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4H+ + 4 e- )} •2× { (6 e- + Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O)} •3 CH3CH2CH2CH2-OH + 3 H2O 3 CH3CH2CH2COOH + 12H+ + 12 e- •12 e- + 2 Cr2O72- + 28 H+ 4 Cr3+ + 14 H2O14
  15. 15. Add the two half-reactions together 3 CH3CH2CH2CH2-OH + 3 H2O + 12e- + 2Cr2O72- + 28 H+ 3 CH3CH2CH2COOH + 12 H+ + 12 e- + 4 Cr3+ + 14 H2O Canceling out the electrons and extra waters, etc. 3 CH3CH2CH2CH2-OH + 2 Cr2O72- + 16 H+ 3 CH3CH2CH2COOH + 4 Cr3+ + 11 H2O15
  16. 16. Spectator ions 3 CH3CH2CH2CH2-OH + 2K2Cr2O7 + 8H2SO4 3 CH3CH2CH2COOH + 2 Cr2(SO4)3 + 11 H2O + 2 K2SO4 Final equation: 3 CH3CH2CH2CH2-OH + 2 K2Cr2O7 + 8 H2SO4 3 CH3CH2CH2COOH + 2 Cr2(SO4)3 + 11 H2O + 2K2SO4 BALANCED16
  17. 17. reduced oxidize17
  18. 18. SUALEHA IQBAL..18

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