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Balancing Redox Equations. Balancing Redox Equations. In balancing redox equations, the # of electrons lost in oxidation (the increase in ox. #) must equal the # of electrons gained in reduction (the decrease in ox. #) There are 2 methods for balancing redox equations:
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Balancing Redox Equations • In balancing redox equations, the # of electrons lost in oxidation (the increase in ox. #) must equal the # of electrons gained in reduction (the decrease in ox. #) • There are 2 methods for balancing redox equations: • Change in Oxidation-Number Method • The Half-Reaction Method
1. Change in Oxidation-Number Method: • based on equal total increases and decreases in oxidation #’s Steps: page 645 in textbook • Write equation and assign oxidation #’s. • Determine which element is oxidized and which is reduced, and determine the change in oxidation # for each. • Connect the atoms that change ox. #’s using a bracket; write the change in ox. # at the midpoint of each bracket. • Choose coefficients that make the total increase in ox. # = the total decrease in ox. #. • Balance the remaining elements by inspection.
S + HNO3 SO2 + NO + H2O Example +4 So, 3 coefficient x +4 = +12 3 4 3 4 2 0 +1 +5 -2 +4 -2 +2 -2 +1 -2 -3 So, 4 coefficient x -3 = -12 Step 1: Oxidation numbers Step 2: Which was oxidized? S. By how much? 0 to +4 = change of +4 +5 to +2 = change of -3 Which was reduced? N. By how much?
If needed, reactions that take place in acidic or basic solutions can be balanced as follows:
Example: Balance the following equation, assuming it takes place in acidic solution. Page 648 in textbook ClO4- + I- Cl- + I2 +1 8 +8 H+ + H2O 4 4 -2 -1 +7 -1 0 Step 1: Oxidation numbers -8 Step 2: Which was oxidized? Iodine, -1 to 0 = +1 Which was reduced? Chlorine, +7 to -1 = -8 Step 5: Balance acid soln with water…
Example: Balance the following equation, assuming it takes place in basic solution. ClO4- + I- Cl- + I2 Cancel H+ by adding OH- to both sides. H+ + OH- = H2O Cancel H2O
2. The Half-Reaction Method: • separate and balance the oxidation and reduction half-reactions. Steps: • Write equation and assign oxidation #’s. • Determine which element is oxidized and which is reduced, and determine the change in oxidation # for each. • Construct unbalanced oxidation and reduction half reactions. • Balance the elements and the charges (by adding electrons as reactants or products) in each half-reaction. • Balance the electron transfer by multiplying the balanced half-reaction by appropriate integers. • Add the resulting half-reaction and eliminate any common terms to obtain the balanced equation.
Example: Balance the following using the half-reaction method: HNO3 + H2S NO + S + H2O
Example: Balance the following using the half-reaction method: HNO3 + H2S NO + S + H2O 2 3 2 3 4 +1 +5 -2 +2 -2 0 -2 +1 -2 +1 Step 1: oxidation numbers Step 3: unbalanced half-rxns x3 S2- S N5+ N2+ + 2 e- 3S2- 3S + 6e- Step 2: Which was oxidized? x2 + 3 e- 2 N5+ + 6e- 2 N2+ Step 4: balance the half – rxns by adding electrons ---------------------------------- 3S2- + 2N5+ 3S + 2N2+ S. -2 to 0 = +2 Which was reduced? Step 6: Add half-rxns and cancel any common terms to obtain a balanced eq. Step 5: balance electron transfer by multiplying by appropriate integers N. +5 to +2 = -3 Now, balance the eq. w/coefficients
If needed, reactions that take place in acidic solutions can be balanced as follows:
If needed, reactions that take place in basic solutions can be balanced as follows:
HOMEWORK: Balance the following using the half-rxn method… In acidic sol’n: a) Cu + NO3- Cu2+ + NO b) Cr2O72- + Cl- Cr3+ + Cl2 c) Pb + PbO2 + H2SO4 PbSO4 In basic sol’n: a) Al + MnO4- MnO2 + Al(OH)4- b) Cl2 Cl- + OCl- c) NO2- + Al NH3 + AlO2-
