Question

An aluminium oxalate complex was prepared by dissolving aluminium metal in potassium hydroxide, followed by reaction with oxalic acid. The resulting complex was analysed for oxalate content by titration against potassium permanganate:

A 0.1105 g sample of the aluminium oxalate complex was dissolved in deionized water (30 cm³) and bench dilute (2 M) sulphuric acid (10 cm³) added. The resulting solution was heated to about 60 ºC and titrated against standardized potassium manganate (VII) solution. 14.35 cm³ of a 0.02 mol dm^-3 KMn0₄ solution was required.

                        (i)        Write out a balanced equation for the overall redox process.

                                                                       

                        (ii)       Use the analytical data to deduce if the prepared complex was K₃Al(C₂O₄)₃.3H₂O or KAl(C₂O₄)₂(H₂O)₂.2H₂O

                        Molar masses /g mol^-1

                        K, 39.00; Al, 26.98; C, 12.01; O, 16.00; H, 1.0

Answer 1

Balancing Redox Equations Using the Oxidation Number Method: oxidation number method,

: Balancing Redox Equations Using the Oxidation Number Technique

Tobalance an equation whether the reaction is a redox reaction or not, follow the steps are

General Steps  

Step 1:  Try to balance the atoms in the equation by inspection, that is, by the standard technique for balancing non-redox equations. (Many equations for redox reactions can be easily balanced by inspection.)

balance the atoms, go to Step 2.

If we unable to balance the atoms, go to Step 3.

Step 2:  Check to be sure that the net charge is the same on both sides of the equation.

If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step 3.

Step 3:  If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and use them to decide whether the reaction is a redox reaction. If it is not redox, return to Step 1 and try again. If it is redox, go to Step 4.

Step 4:  Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.

Step 5:  Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (a ratio that makes the number of electrons lost equal to the number of electrons gained).

Step 6:  Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. (These coefficients are usually placed in front of the formulas on the reactant side of the arrow.)

Step 7:  Balance the rest of the equation by inspection

FOR EXAMPLE sodiumoxalate withH2SO4

KMnO4+Na2C2O4+H2SO4→K2SO4+Na2SO4+MnSO4+CO2+H2O

Step 1. The oxidation numbers are:

Left hand side: K=+1;Mn=+7;O=−2;Na=+1;C=+3:H=+1;S=+6

Right hand side: K=+1;S=+6;O=−2;Na=+1;Mn=+2;C=+4;H=+1

Step 2. The changes in oxidation number are:

Mn: +7 → +2; Change = -5
C: +3 → +4; Change = +1

Step 3. Equalize the changes in oxidation number.

You need 1 atom of Mn for every 5 atoms of C or 2 atoms of Mn for every 10 atoms of C

. This gives us total changes of +10 and −10

Step 4. Insert coefficients to get these numbers.

2KMnO4+5Na2C2O4+H2SO4→K2SO4+Na2SO4+2MnSO4+10CO2+H2O

Step 5. Balance K.

2KMnO4+5Na2C2O4+H2SO4→1K2SO4+Na2SO4+2MnSO4+10CO2+H2O

Step 6. Balance Na.

2KMnO4+5Na2C2O4+H2SO4→1K2SO4+5Na2SO4+2MnSO4+10CO2+H2O

Step 7. Balance S.

2KMnO4+5Na2C2O4+8H2SO4→1K2SO4+5Na2SO4+2MnSO4+10CO2+H2O

Step 8. Balance O.

2KMnO4+5Na2C2O4+8H2SO4→1K2SO4+5Na2SO4+2MnSO4+10CO2+8H2O

Every substance now has a coefficient.

Step 9. Check that all atoms balance.

Left hand side: 2K; 2 Mn; 60 O; 10 Na; 10 C; 16 H; 8 S
Right hand side: 2 K; 8 S; 60 O; 10 Na; 2 Mn; 10 C; 16 H

The balanced equation is

2KMnO4+5Na2C2O4+8H2SO4→K2SO4+5Na2SO4+MnSO4+10CO2+8H2O