* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 3.9-10 Redox titrations
Survey
Document related concepts
Electron configuration wikipedia , lookup
Chemical bond wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Stability constants of complexes wikipedia , lookup
Equilibrium chemistry wikipedia , lookup
Ionic compound wikipedia , lookup
Van der Waals equation wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Heat equation wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Equation of state wikipedia , lookup
Detailed balance wikipedia , lookup
Transcript
Redox titrations Textbook reference: p 218-222 Redox titrations • Carry out redox titrations and carry out structured calculations involving MnO4• Understand how to carry out redox titrations and carry out structured calculations involving MnO4• Perform non-structured titration calculations, based on experimental results • Carry out structured titration calculations based on experimental results Redox titrations • These are based on the same principles as acid-base titrations but involves the transfer of electrons rather than the transfer of protons • You need an indicator species which is any chemical that will change colour when all of the oxidising agent has been used up Self-indicating reactions • Some redox titrations do not need an indicator like acid-base titrations do • The strong oxidiser potassium manganate (VII) ions are a good example MnO4- (aq) Mn2+ (aq) Purple Almost colourless Can you think of any other oxidising agents that change colour as they change oxidation state? Iron in redox reactions • Whenever iron is used with an oxidising agent it will most probably be as the 2+ ion, e.g. when metallic iron is dissolved in sulfuric acid BALANCING REDOX HALF EQUATIONS 1 2 3 4 Work out formulae of the species before and after the change; balance if required Work out oxidation state of the element before and after the change Add electrons to one side of the equation so that the oxidation states balance If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 Step 1 MnO4¯ being reduced to Mn2+ in acidic solution MnO4¯ ———> Mn2+ No need to balance Mn; equal numbers BALANCING REDOX HALF EQUATIONS 1 2 3 4 Work out formulae of the species before and after the change; balance if required Work out oxidation state of the element before and after the change Add electrons to one side of the equation so that the oxidation states balance If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 Step 1 Step 2 MnO4¯ being reduced to Mn2+ in acidic solution MnO4¯ ———> +7 Mn2+ +2 Overall charge on MnO4¯ is -1; sum of the OS’s of all atoms must add up to -1 Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8 To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1] BALANCING REDOX HALF EQUATIONS 1 2 3 4 Work out formulae of the species before and after the change; balance if required Work out oxidation state of the element before and after the change Add electrons to one side of the equation so that the oxidation states balance If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 Step 1 Step 2 Step 3 MnO4¯ being reduced to Mn2+ in acidic solution MnO4¯ ———> Mn2+ +7 +2 MnO4¯ + 5e¯ ———> Mn2+ The oxidation states on either side are different; +7 —> +2 (REDUCTION) To balance; add 5 negative charges to the LHS [+7 + (5 x -1) = +2] You must ADD 5 ELECTRONS to the LHS of the equation BALANCING REDOX HALF EQUATIONS 1 2 3 4 Work out formulae of the species before and after the change; balance if required Work out oxidation state of the element before and after the change Add electrons to one side of the equation so that the oxidation states balance If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 Step 1 Step 2 Step 3 Step 4 MnO4¯ being reduced to Mn2+ in acidic solution MnO4¯ ———> Mn2+ +7 +2 MnO4¯ + 5e¯ ———> Mn2+ MnO4¯ + 5e¯ + 8H+ ———> Mn2+ Total charges on either side are not equal; LHS = 1- and 5- = 6RHS = 2+ Balance them by adding 8 positive charges to the LHS [ 6- + (8 x 1+) = 2+ ] You must ADD 8 PROTONS (H+ ions) to the LHS of the equation BALANCING REDOX HALF EQUATIONS 1 2 3 4 Work out formulae of the species before and after the change; balance if required Work out oxidation state of the element before and after the change Add electrons to one side of the equation so that the oxidation states balance If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one side Example 2 Step 1 Step 2 Step 3 Step 4 Step 5 MnO4¯ being reduced to Mn2+ in acidic solution MnO4¯ ———> Mn2+ +7 +2 MnO4¯ + 5e¯ ———> Mn2+ MnO4¯ + 5e¯ + 8H+ ———> Mn2+ MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O Everything balances apart from oxygen and hydrogen O LHS = 4 H now balanced RHS = 0 LHS = 8 You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced RHS = 0 5.3 Exercise 1 BALANCING REDOX HALF EQUATIONS Q. Balance the following half equations... Na —> Na+ Fe2+ —> Fe3+ I2 —> I¯ C2O42- —> CO2 H2O2 1 2 3 4 5 —> O2 H2O2 —> H2O NO3- —> NO NO3- —> NO2 SO42- —> SO2 REMINDER Work out the formula of the species before and after the change; balance if required Work out the oxidation state of the element before and after the change Add electrons to one side of the equation so that the oxidation states balance If the charges on all the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges If the equation still doesn’t balance, add sufficient water molecules to one side BALANCING REDOX HALF EQUATIONS Q. Balance the following half equations... Na —> Na+ + e- Fe2+ —> Fe3+ + e- + 2e- I2 + 2e- C2O42H2O2 —> 2I¯ —> 2CO2 —> O2 + 2H+ + 2e- H2O2 + 2H+ + 2e- —> 2H2O NO3- + 4H+ + 3e- —> NO + 2H2O NO3- + 2H+ + e- —> NO2 + H 2O SO42- + 4H+ + 2e- —> SO2 + 2H2O COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Step 2 Step 3 Step 4 Write out the two half equations Multiply the equations so that the number of electrons in each is the same Add the two equations and cancel out the electrons on either side If necessary, cancel any other species which appear on both sides Worked example MnO4-(aq) + 8H+(aq) + 5eFe2+(aq) Mn2+(aq) + 4H2O(l) Fe3+(aq) + e- 1. Construct the fully balanced redox ionic equation for the manganate(VII) ion oxidising the iron(II) ion. 2. 24.3 cm3 of 0.02 mol dm-3 KMnO4 reacted with 20.0 cm3 of an iron(II) solution. 3. Calculate the concentration of the iron(II) ion. 4. How do recognise the end-point in the titration? 5. Calculate the percentage of iron in a sample of steel wire if 1.51 g of the wire was dissolved in excess of dilute sulphuric acid and the solution made up to 250 cm3 in a standard graduated flask. 25.0 cm3 of this solution was pipetted into a conical flask and needed 25.45 cm3 of 0.02 mol dm-3 KMnO4 for complete oxidation. • Calculate the percentage of iron in a sample of steel wire if 1.51 g of the wire was dissolved in excess of dilute sulphuric acid and the solution made up to 250 cm3 in a standard graduated flask. 25.0 cm3 of this solution was pipetted into a conical flask and needed 25.45 cm3 of 0.02 mol dm-3 KMnO4 for complete oxidation.