Redox Reaction
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Redox Reaction

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Redox Reaction Presentation Transcript

  • 1. Oxidation- Reduction T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
  • 2. Introduction • Redox reactions include reactions which involve the loss or gain of electrons. • The reactant giving away (donating) electrons is called the reducing agent (which is oxidized) • The reactant taking (accepting) electrons is called the oxidizing agent (which is reduced) • Both oxidation and reduction happen simultaneously, however each is considered separately using ion- electron equations. © iTutor. 2000-2013. All Rights Reserved
  • 3. Oxidation-Reduction  Oxidation: Loss of electrons.  Reduction: Gain of electrons.  Reductant: Species that loses electrons.  Oxidant: Species that gains electrons. Oxidation Reduction Valence: the electrical charge an atom would acquire if it formed ions in aqueous solution. © iTutor. 2000-2013. All Rights Reserved
  • 4. Points to Remember Oxidizing agent Reducing agent Is itself reduced Is itself oxidized Gains electrons Loses electrons Causes oxidation Causes reduction © iTutor. 2000-2013. All Rights Reserved
  • 5. VARIABLE VALENCE ELEMENTS  Sulfur: SO4 2-(+6), SO3 2-(+4), S(0), FeS2(-1), H2S(-2)  Carbon: CO2(+4), C(0), CH4(-4)  Nitrogen: NO3 -(+5), NO2 -(+3), NO(+2), N2O(+1), N2(0), NH3(-3)  Iron: Fe2O3(+3), FeO(+2), Fe(0)  Manganese: MnO4 -(+7), MnO2(+4), Mn2O3(+3), MnO(+2), Mn(0)  Copper: CuO(+2), Cu2O(+1), Cu(0)  Tin: SnO2(+4), Sn2+(+2), Sn(0)  Uranium: UO2 2+(+6), UO2(+4), U(0)  Arsenic: H3AsO4 0(+5), H3AsO3 0(+3), As(0), AsH3(-1)  Chromium: CrO4 2-(+6), Cr2O3(+3), Cr(0)  Gold: AuCl4 -(+3), Au(CN)2 -(+1), Au(0) © iTutor. 2000-2013. All Rights Reserved
  • 6. Balancing Redox equations (pH acid)  Identify the atoms that are oxidized and reduced, using ox. no.  Write the half-reactions for oxidation and reduction processes.  Balance mass, so that the number of atoms of each element oxidized/reduced is the same on both sides. If there is any O atoms present, balance them adding H2O. If there is H atoms present, balance them adding H+.  Balance charges with e- (electron).  Multiply the half-equations so that the number of e- lost in one equation is equal to the e- gained in the other.  Add the two half-equation together and cancel out any substances that appear on both side. © iTutor. 2000-2013. All Rights Reserved
  • 7. Cl- + MnO4 -  Mn2+ + Cl2 (Acid Medium) Identify atoms oxidised and reduced, using ox. no. Cl : -1 Mn : +7 O: -2  Mn2+: +2 Cl: 0 Cl oxidises (goes from -1 to 0) Mn reduces (from +7 to +2) Write the half-reactions. Balance mass, if necessary using H2O (to add O) and/or H+ (to add H) 2Cl-  Cl2 Oxidation (reducing agent) 8H+ + MnO4 -  Mn2+ + 4H2O Reduction (oxidising agent) © iTutor. 2000-2013. All Rights Reserved
  • 8. Balance the charges using electrons (e-). Each e- adds a -1 2Cl-  Cl2 + 2e- -2 = 0 -2 8H+ + MnO4 - + 5e-  Mn2+ + 4H2O +8 -1 -5 = +2 +0 Multiply the half-equations so that e- lost in one = e- gained in the other 5x(2Cl-  Cl2 + 2e-) 2x(8H+ + MnO4 - + 5e-  Mn2 + + 4H2O) Cl- + MnO4 -  Mn2+ + Cl2 (Acid Medium) © iTutor. 2000-2013. All Rights Reserved
  • 9. Add the two half-equation together and cancel out any substances that appear on both sides 10Cl-  5Cl2 + 10e- 16H+ + 2MnO4 - + 10e-  2Mn2+ + 8H2O) 10Cl- + 16H+ + 2MnO4 - + 10e-  5Cl2 + 10e- + 2Mn2+ + 8H2O Answer: 10Cl- + 16H+ + 2MnO4 -  5Cl2 + 2Mn2+ + 8H2O + Cl- + MnO4 -  Mn2+ + Cl2 (Acid Medium) © iTutor. 2000-2013. All Rights Reserved
  • 10. Balancing redox equations (pH basic)  Identify the atoms that are oxidised and reduced, using ox. no.  Write the half reactions for oxidation and reduction processes  Balance mass, so that the number of atoms of each element oxidised/reduced is the same on both sides If there is any O atoms present, balance them adding OH- If there is H atoms present, balance them adding H2O  Balance charges with e-  Multiply the half-equations so that the number of e- lost in one equation is equal to the e- gained in the other  Add the two half-equation together,cancel out any substances that appear on both side. © iTutor. 2000-2013. All Rights Reserved
  • 11. CrO4 2- + Fe2+  Cr3+ + Fe3+ (basic pH) Identify the atoms oxidised and reduced, using ox. no. Fe : +2 Cr : +6 O: -2 Cr3+: +3 Fe3+: +3 Fe oxidises (goes from +2 to +3) Cr reduces (goes from +6 to +3) Write the half-reactions. Balance mass, if necessary using OH- (to add O, double the number needed) and/or H2O (to add H) Fe2+  Fe3+ Oxidation (reducing agent) 4H2O + CrO4 -2  Cr3+ + 8OH- Reduction (oxidising agent) © iTutor. 2000-2013. All Rights Reserved
  • 12. CrO4 2- + Fe2+  Cr3+ + Fe3+ (basic pH) Balance the charges using electrons (e-). Each e- adds a -1 Fe2+  Fe3+ + 1 e- (+2 = +3 – 1) 4H2O + CrO4 -2 + 3 e-  Cr3+ + 8OH-(0 -2 -3 = +3 -8) Multiply the half-equations so that e- lost in one = e- gained in the other 3x(Fe2+  Fe3+ + 1e-) 1x(4H2O + CrO4 -2 + 3e-  Cr3+ + 8OH-) © iTutor. 2000-2013. All Rights Reserved
  • 13. CrO4 2- + Fe2+  Cr3+ + Fe3+ (basic pH) Add the two half-equation together and cancel out Any substances that appear on both side 3Fe2+  3Fe3+ + 3e- 4H2O + CrO4 -2 + 3e-  Cr3+ + 8OH- 3Fe2+ + 4H2O + CrO4 -2 + 3e-  3Fe3+ + 3e- +Cr3+ + 8OH- Answer: 3Fe2+ + 4H2O + CrO4 -2  3Fe3+ +Cr3+ + 8OH- + © iTutor. 2000-2013. All Rights Reserved
  • 14. The End Call us for more Information: www.iTutor.com Visit 1-855-694-8886