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9.2.1: Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction Step 1: Identify the oxidation states of the.

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Presentation on theme: "9.2.1: Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction Step 1: Identify the oxidation states of the."— Presentation transcript:

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2 9.2.1: Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction Step 1: Identify the oxidation states of the species on either side of the reaction. Step 2: Identify the oxidation half reaction by identifying which reactant undergoes oxidation. Identify the reduction half reaction by identifying which reactant undergoes reduction. Step 3: Deduce the number of electrons transferred and produce half-equations.

3 Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. Cl 2 + KI  I 2 + KCl This is the unbalanced skeleton equation. Redox half-reactions can be used to balance complex reaction equations

4 Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. Step 1: Identify the oxidation states of the species on either side of the reaction. Cl K +1 I -1  I K +1 Cl -1

5 Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. Step 2: Identify the OXidation half reaction by identifying which reactant undergoes oxidation. Identify the REDuction half reaction by identifying which reactant undergoes oxidation. Cl K +1 I -1  I K +1 Cl -1 Cl 2 is reduced to Cl - I -1 is oxidised to I 2 K + does not change = spectator OX RED

6 Example: In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. Step 3: Deduce the number of electrons transferred and produce half-equations. O XIDATION H ALF -R EACTION (Oxidation is Loss of Electrons) Electrons are removed from the reactant (appear as product) Balance number of atoms 2I -1  I e - CHARGE IS CONSERVED RED UCTION H ALF -R EACTION (Reduction is Gain of Electrons) Electrons are added to a reactant. Cl e -  2Cl -1

7 9.2.2: Deduce redox equations using half-equations. This means balance chemical equations using electrons in half-reactions H + and H 2 O should be used where necessary to balance half -equations in acid solution. The balancing of equations for reactions in alkaline solution will not be assessed.

8 Balancing Redox in Acid Add these steps to balance OXYGEN and HYDROGEN atoms in redox half-reactions Step 1: Balance OXYGEN by adding WATER (H 2 O) Step 2: Balance the HYDROGEN by adding H + ions

9 Oxidation of Ethanol using Acidified Dichromate CH 3 CH 2 OH + Cr 2 O 7 2-  CH 3 COOH + Cr 3+ Rewritten: C 2 H 6 O + Cr 2 O 7 2-  C 2 H 4 O 2 + Cr 3+ Assign oxidation numbers (H + and O 2- stay constant here) (C 2+ ) 2 H 6 O + (Cr 6+ ) 2 O 7 2-  (C 0 ) 2 H 4 O 2 + Cr 3+ Carbon in ethanol gets oxidised from 2+ to 0 Chromium in dichromate gets reduced from 6+ to 3+

10 Oxidation Half-Reaction in Acid O XIDATION H ALF -R EACTION (2e - per C = 4e - ) C 2 H 6 O  C 2 H 4 O 2 + 4e - Add H 2 O to balance oxygens C 2 H 6 O + H 2 O  C 2 H 4 O 2 + 4e - Add H + to balance hydrogens C 2 H 6 O + H 2 O  C 2 H 4 O 2 + 4e - + 4H + net charge on each side is balanced (0) + (0) = (0) + (-4) + (+4) If charge is not balanced, then it is wrong.

11 Reduction Half-Reactions in Acid R EDUCTION H ALF -R EACTION (3e - per Cr = 6e - ) Cr 2 O e -  2Cr 3+ Add H 2 O to balance oxygens Cr 2 O e -  2Cr H 2 O Add H + to balance hydrogens Cr 2 O H + + 6e -  2Cr H 2 O net charge on each side is balanced (-2) + (+14) + (-6) = (+6)+ (0) If charge is not balanced, then it is wrong.

12 Combine Half-Reactions O XIDATION : C 2 H 6 O + H 2 O  C 2 H 4 O 2 + 4e - + 4H + R EDUCTION : Cr 2 O H + + 6e -  2Cr H 2 O Multiply each to balance electrons transferred [C 2 H 6 O + H 2 O  C 2 H 4 O 2 + 4e - + 4H + ] x 3 [Cr 2 O H + + 6e -  2Cr H 2 O] x 2 3C 2 H 6 O + 3H 2 O  3C 2 H 4 O e H + 2Cr 2 O H e -  4Cr H 2 O _______________________________________________________ 3C 2 H 6 O + 3H 2 O + 2Cr 2 O H +  3C 2 H 4 O H + + 4Cr H 2 O 3C 2 H 6 O + 2Cr 2 O H +  3C 2 H 4 O 2 + 4Cr H 2 O Don’t forget to check net charge


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