Analysing Oxidants and Reductants

What is a redox reaction?• The reactant that loses electrons is

oxidisedMnemonic devices • OIL RIG

• O(xidation)• I(s)• L(oss of electrons)

• R(eduction)• I(s)• G(ain of electrons)

• The reactant that gains electrons is reduced

• LEO the lion says GER• L(oss of)• E(lectrons is)• O(xidation)

• G(ain of • E(lectrons is)• R(eduction)

Oxidation and reduction occur simultaneously• Through the loss of electrons the reactant that is oxidised causes the

reduction of the other reactant® Therefore it is a reducing agent or reductant

• Through the gain of electrons the reactant that is reduced causes the oxidation of the other reactant

® Therefore it is an oxidising agent or oxidant

2Fe2O3(s) + 3C(s) ® 4Fe(l) + 3CO2(g)

reduction

oxidationoxidant reductant

Identifying redox reactionsOxidation number rules• Free elements have an oxidation number of zero ()• Ionic compounds composed of simple ions have an oxidation number

equal to the charge on the ion () • Oxygen is usually -2, except in a peroxide (H2O2)• Hydrogen is usually +1, except in a metal hydride (NaH)• The sum of the oxidation numbers in a compound equals the charge

on the compound (zero, if a neutral compound)• The most electronegative element in a compound has the negative

oxidation number (F>O>Cl>N>other elements)

0

+1 -1

+1 -1

+1 -1

Applying oxidation number rules

Reactants• Fe2O3: O has an oxidation number of -2, so 2x + (3)(-2) = 0

\ Fe in Fe2O3has an oxidation number of +3• C: C as a free element has an oxidation number of 0Products• Fe: Fe as a free element has an oxidation number of 0• CO2: O has an oxidation number of -2, so x + (2)(-2) = 0

\ C in CO2 has an oxidation number of +4 Iron is reduced (+3 ® 0) and carbon is oxidised (0 ® +4)

2Fe2O3(s) + 3C(s) ® 4Fe(l) + 3CO2(g)

Writing and balancing redox half reactionsThe “half reactions” only include the oxidation or reduction reactions• Oxidation of magnesium

Mg(s) ® Mg2+(aq)

• Reduction of nitric acidNO3

-(aq) ® N2O(g)

Neither reaction is balanced for charge or mass• Mg(s) ® Mg2+(aq) + 2e- (mass is balanced; add electrons to balance charge)

• 2NO3-(aq) ® N2O(g) (balance mass except O and H)

• 2NO3-(aq) ® N2O(g) + 5H2O(l) (balance oxygen by adding water)

• 2NO3-(aq) + 10H+(aq) ® N2O(g) + 5H2O(l) (balance hydrogen by adding H+)

• 2NO3-(aq) + 10H+(aq) 8e- ® N2O(g) + 5H2O(l) (balance charge by added e-)

0 2+

5+ 2- 1+ 2-

Volumetric analysis using redox titration (1)Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H3PO4/H2SO4 mixture to reduce all of the iron to Fe2+ ions. The solution is then titrated with 0.01625 M K2Cr2O7, producing Fe3+ and Cr3+ ions in acidic solution. The titration requires 32.26 mL of K2Cr2O7 for 1.2765 g of the sample.

Where do we start?• Identify key information:

® All iron is in the reduced state (Fe2+)® The solution is acidic (important to note when balancing half reactions)® Fe2+ is oxidised to Fe3+ (oxidation number increased from +2 to +3)® Chromium in Cr2O7

2- is reduced to Cr3+ (oxidation number reduced from +6 to +3) ® PO4

3-, SO42- and K+ are all spectator ions

® Answer should be in 4 significant digits

Volumetric analysis using redox titration (2)Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H3PO4/H2SO4 mixture to reduce all of the iron to Fe2+ ions. The solution is then titrated with 0.01625 M K2Cr2O7, producing Fe3+ and Cr3+ ions in acidic solution. The titration requires 32.26 mL of K2Cr2O7 for 1.2765 g of the sample.

Write out balanced half reactionsOxidation• Fe2+ ® Fe3+ + e-

Reduction• Cr2O7

2- ® 2Cr3+ (balance Cr)• Cr2O7

2- ® 2Cr3+ + 7H2O (balance O)• Cr2O7

2- + 14H+ ® 2Cr3+ + 7H2O (balance H; remember it is in an acidic solution)• Cr2O7

2- + 14H+ + 6e- ® 2Cr3+ + 7H2O (balance charge)

Volumetric analysis using redox titration (3)Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H3PO4/H2SO4 mixture to reduce all of the iron to Fe2+ ions. The solution is then titrated with 0.01625 M K2Cr2O7, producing Fe3+ and Cr3+ ions in acidic solution. The titration requires 32.26 mL of K2Cr2O7 for 1.2765 g of the sample.

Write balanced full reaction including statesHalf reactions• 6Fe2+ ® 6Fe3+ + 6e-

• Cr2O72- + 14H+ + 6e- ® 2Cr3+ + 7H2O

Full reaction• Cr2O7

2-(aq) + 6Fe2+(aq) + 14H+(aq) ® 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

Volumetric analysis using redox titration (4)Potassium dichromate is used to titrate a sample containing an unknown percentage of iron. The sample is dissolved in H3PO4/H2SO4 mixture to reduce all of the iron to Fe2+ ions. The solution is then titrated with 0.01625 M K2Cr2O7, producing Fe3+ and Cr3+ ions in acidic solution. The titration requires 32.26 mL of K2Cr2O7 for 1.2765 g of the sample.

Calculate percentage of iron in sample• Cr2O7

2-(aq) + 6Fe2+(aq) + 14H+(aq) ® 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) • n(Cr2O7

2-) = (0.01625 molL-1)(0.03226 L) = 0.000524225 mol• n(Cr2O7

2-):n(Fe(II)) = 1:6• n(Fe(II)) = 6 × 0.000524225 mol = 0.00314535 mol• m(Fe(II)) = (0.00314535 mol)(55.845 gmol-1) = 0.17565 g• Percentage of iron in sample = (0.17565 g/1.2765 g) × 100 = 13.76%