Analysing Oxidants and Reductants

What is a redox reaction? The reactant that loses electrons is oxidised Mnemonic 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 2Fe 2 O 3 (s) + 3C (s)  4Fe (l) + 3CO 2 (g) reduction oxidation oxidant reductant

Identifying redox reactions

Applying oxidation number rules Reactants Fe 2 O 3 :O has an oxidation number of -2, so 2 x + (3)(-2) = 0  Fe in Fe 2 O 3 has an oxidation number of +3 C:C as a free element has an oxidation number of 0 Products Fe:Fe as a free element has an oxidation number of 0 CO 2 :O has an oxidation number of -2, so x + (2)(-2) = 0  C in CO 2 has an oxidation number of +4  Iron is reduced (+3  0) and carbon is oxidised (0  +4) 2Fe 2 O 3 (s) + 3C (s)  4Fe (l) + 3CO 2 (g)

Writing and balancing redox half reactions The “half reactions” only include the oxidation or reduction reactions Oxidation of magnesium Mg (s)  Mg 2+ (aq) Reduction of nitric acid NO 3 - (aq)  N 2 O (g) Neither reaction is balanced for charge or mass Mg (s)  Mg 2+ (aq) + 2e - (mass is balanced; add electrons to balance charge) 2NO 3 - (aq)  N 2 O (g) (balance mass except O and H) 2NO 3 - (aq)  N 2 O (g) + 5H 2 O (l) (balance oxygen by adding water) 2NO 3 - (aq) + 10H + (aq)  N 2 O (g) + 5H 2 O (l) (balance hydrogen by adding H + ) 2NO 3 - (aq) + 10H + (aq) 8e -  N 2 O (g) + 5H 2 O (l) (balance charge by added e - )

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 H 3 PO 4 /H 2 SO 4 mixture to reduce all of the iron to Fe 2+ ions. The solution is then titrated with M K 2 Cr 2 O 7, producing Fe 3+ and Cr 3+ ions in acidic solution. The titration requires mL of K 2 Cr 2 O 7 for g of the sample. Where do we start? Identify key information:  All iron is in the reduced state (Fe 2+ )  The solution is acidic (important to note when balancing half reactions)  Fe 2+ is oxidised to Fe 3+ (oxidation number increased from +2 to +3)  Chromium in Cr 2 O 7 2- is reduced to Cr 3+ (oxidation number reduced from +6 to +3)  PO 4 3-, SO 4 2- 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 H 3 PO 4 /H 2 SO 4 mixture to reduce all of the iron to Fe 2+ ions. The solution is then titrated with M K 2 Cr 2 O 7, producing Fe 3+ and Cr 3+ ions in acidic solution. The titration requires mL of K 2 Cr 2 O 7 for g of the sample. Write out balanced half reactions Oxidation Fe 2+  Fe 3+ + e - Reduction Cr 2 O 7 2-  2Cr 3+ (balance Cr) Cr 2 O 7 2-  2Cr H 2 O (balance O) Cr 2 O H +  2Cr H 2 O (balance H; remember it is in an acidic solution) Cr 2 O H + + 6e -  2Cr H 2 O (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 H 3 PO 4 /H 2 SO 4 mixture to reduce all of the iron to Fe 2+ ions. The solution is then titrated with M K 2 Cr 2 O 7, producing Fe 3+ and Cr 3+ ions in acidic solution. The titration requires mL of K 2 Cr 2 O 7 for g of the sample. Write balanced full reaction including states Half reactions 6Fe 2+  6Fe e - Cr 2 O H + + 6e -  2Cr H 2 O Full reaction Cr 2 O 7 2- (aq) + 6Fe 2+ (aq) + 14H + (aq)  2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2 O (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 H 3 PO 4 /H 2 SO 4 mixture to reduce all of the iron to Fe 2+ ions. The solution is then titrated with M K 2 Cr 2 O 7, producing Fe 3+ and Cr 3+ ions in acidic solution. The titration requires mL of K 2 Cr 2 O 7 for g of the sample. Calculate percentage of iron in sample Cr 2 O 7 2- (aq) + 6Fe 2+ (aq) + 14H + (aq)  2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2 O (l) n(Cr 2 O 7 2- ) = ( mol  L -1 )( L) = mol n(Cr 2 O 7 2- ):n(Fe(II)) = 1:6 n(Fe(II)) = 6 × mol = mol m(Fe(II)) = ( mol)( g  mol -1 ) = g Percentage of iron in sample = ( g/ g) × 100 = 13.76%