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# 2 Redox Titrations

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### 2 Redox Titrations

1. 1. Redox Titrations
2. 2. <ul><li>Oxidation-reduction reactions involve a transfer of electrons. The oxidising agent accepts electrons and the reducing agent gives electrons. In working out the equation for a redox reaction we can use the half-equation method. </li></ul><ul><li>In this method we use the half-equation for the oxidising agent and the half-equation for the reducing agent then add them together. </li></ul>
3. 3. <ul><li>Examples of half-reaction equations </li></ul><ul><li>Iron(III) salts are reduced to iron(II) salts. </li></ul><ul><li>The half equation is </li></ul><ul><li>Fe 3+  Fe 2+ </li></ul><ul><li>For the equation to balance the charge on the RHS must equal the charge on the LHS. This can be accomplished by inserting an electron on the LHS </li></ul><ul><li>Fe 3+ + e -  Fe 2+ </li></ul>
4. 4. <ul><li>When chlorine acts as an oxidising agent it is reduced to Cl - ions </li></ul><ul><li>Cl 2  2Cl - </li></ul><ul><li>To obtain a balanced equation 2 electrons muct be inserted on the LHS </li></ul><ul><li>Cl 2 + 2e -  2Cl - </li></ul>
5. 5. <ul><li>Potassium mangante(VII) is an oxidising agent. In acid solutions it is reduced to a manganese(II) salt </li></ul><ul><li>MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O </li></ul><ul><li>Potassium dichromate(VI) is also an oxidising agent. It is reduced to a chromium(III) salt </li></ul><ul><li>Cr 2 O 7 2- + 14H + + 6e -  2Cr 3+ + 7H 2 O </li></ul>
6. 6. <ul><li>Using half equations to obtain the equation for the reaction </li></ul><ul><li>Find the equation for the reaction of iron(III) with chloride ions </li></ul><ul><li>The 2 half reactions needed are </li></ul><ul><li>1 Fe 3+ + e -  Fe 2+ </li></ul><ul><li>2 Cl 2 + 2e -  2Cl - </li></ul><ul><li>We need iron(III) and chloride as reactants so equation 2 must be reversed </li></ul><ul><li>3 2Cl -  Cl 2 + 2e - </li></ul>
7. 7. <ul><li>Now we must balance the electrons, then add the half equations together to obtain the full equation. To balance the electrons we multiply equation 1 by 2 </li></ul><ul><li>Fe 3+ + e -  Fe 2+ x2 </li></ul><ul><li>4 2Fe 3+ + 2e -  2Fe 2+ </li></ul><ul><li>3 2Cl -  Cl 2 + 2e - </li></ul><ul><li>2Fe 3+ + 2Cl -  2Fe 2+ + Cl 2 </li></ul><ul><li>There must be the same no of electrons on each side which then cancel. </li></ul>
8. 8. <ul><li>Construct an equation for the reaction of potassium manganate(VII) with ethanedioate ion </li></ul><ul><li>Ethanedioate ion is C 2 O 4 2- and the half equation is </li></ul><ul><li>C 2 O 4 2-  2CO 2 + 2e - </li></ul>2MnO 4 - + 16H + + 5C 2 O 4 2-  2Mn 2+ + 8H 2 O + 10CO 2
9. 9. <ul><li>Now we can do a titration calculation using this equation! </li></ul><ul><li>A 25.0ml portion of sodium ethanedioate solution of concentration 0.200 mol/L is warmed and titrated against a solution of potassium manganate(VII). If 17.2ml of potassium manganate(VII) is required what is its concentration? </li></ul>
10. 10. <ul><li>No of moles ethanedioate in 25.0ml of solution of conc 0.200 mol/L = </li></ul><ul><li>0.200 x 25 = 0.005000 mols </li></ul><ul><li>1000 </li></ul><ul><li>Ratio of ethanedioate to manganate(VII) = 2:5 </li></ul><ul><li> no of mols of manganate(VII) in 17.2ml = </li></ul><ul><li>0.005000 x 2 = 0.002000 mols </li></ul><ul><li>5 </li></ul><ul><li> conc of manganate(VII) = 0.002000 x 1000 </li></ul><ul><li>17.2 </li></ul><ul><li>= 0.116 mol/L </li></ul>
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