Lecture 5: Electrochemistry Lecture 5 Topic Chapter 20 1. Redox agents & half-equations Reducing & oxidizing agents 20.1 Solving redox by half-equation 20.2 Steps! 2. Voltaic cells are redox reactions 20.3 Separate “half-cells” 3. Batteries Electromotive force 20.4 Batteries & Calculating Ecell 20.7 Fuel Cells 4. Corrosion & Electrolysis Corrosion 20.8 Electrolysis 20.9

Redox agents & half-equations Reduction agents are oxidized. Oxidation agents are reduced. Half-equations account for electrons & ‘co-factors’. The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made. This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory. While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

Redox terminology Oxidizing agent (oxidant) the substance that removes e- from the oxidized compound The oxidizing agent is reduced during redox reactions. Reducing agent (reductant) the substance that donates e- to the compound being reduced The reducing agent is oxidized during redox reactions. Zn (solid) + 2H+ --> Zn+2 + H2 (gas) 0 +1 +2 0 oxidation numbers reducing agent oxidizing agent Zn = oxidized Zn becomes + charged, thus lost e- H = reduced H loses + charged, thus has gained e- p. 828-9

Identifying redox agents The rechargeable nickel-cadmium (nicad) battery uses this redox reaction to generate electricity: Cd(s) + NiO2(s) + 2H2O(l) --> Cd(OH)2(s) + Ni(OH)2(s) 0 +4/-2 +1/-2 +2/-2/+1 +2/-2/+1 a) Identify the substances being reduced & oxidized. increasing ox# -> oxidized Cd decreasing ox# -> reduced Ni b) Identify the oxidizing & reducing agents. - thus = reducing agent increasing ox# -> oxidized Cd decreasing ox# -> reduced Ni - thus = oxidizing agent Identify the oxidizing & reducing agents in this redox equation: 2H2O(l) + Al(s) + MnO4-(aq) --> Al(OH)4-(aq) + MnO2(s) +1/-2 0 +7/-2 +3/-2/+1 +4/-2 increasing ox# for Al -> Al is oxidized & is the reducing agent decreasing ox# for Mn -> Mn is reduced & is the oxidizing agent p. 828 - 9

Balancing redox half-equations Usually balancing an equation means getting the same number of atoms of each element on each side of the equation. But, for redox reactions: the number of e- lost and gained on each side must also be balanced. To make this ‘easier’ we think of redox reactions as pairs of half-reactions written as half-equations. Sn+2 + 2Fe+3 --> Sn+4 + 2Fe+2 oxidation: Sn+2 --> Sn+4 + 2e- reduction: 2Fe+3 + 2e- --> 2Fe+2 If e- cancel out and the 2 half-reactions are added they sum to the overall equation. Notice that the e- are added to opposite sides of the 2 half-equations. p. 830 - 3

Steps for balancing half-equations Why use this method? Isn’t it more complicated than simply assigning oxidation numbers? Yes, half-equations appear complicated, but redox reactions are often complex & involve the solvent that the compounds are suspended in. Solvents can participate in the redox reaction itself, & need to accounted for. Steps to follow for acidic solutions: 1. Rewrite the equation in half-reactions, one for each reactant 2. Balance each half-equation for mass starting with non-H and non-O elements. 3. Balance for O atoms by adding H2O where needed. 4. Balance for H atoms by adding H+1 ions where needed.. 5. Balance for charge by adding e- to one side. [Note that e- should be added to opposite sides of the two half-equations.] 6. If necessary, multiply half-equations by factors so that the same number of e- are added or removed in each half-equation. Add the two half-equations, canceling out items that are the same on both sides. [Note that all e- should cancel.] 8. Double check to see that this final equation balances. p. 830 - 3

Example: acidic aqueous solution Consider the reaction between permanganate ion (MnO4-) and oxalate ion (C2O4-2) in an acidic solution. Color changes & a gas is formed. MnO4-1 + C2O4-2 --> Mn+2 + CO2 (gas) H+1 is involved. MnO4-1  Mn+2 C2O4-2  CO2 MnO4-1  Mn+2 + 4H2O C2O4-2  2CO2 -2 8H+1 + MnO4-1  Mn+2 + 4H2O +7 +2 5e- + 8H+1 + MnO4-1  Mn+2 + 4H2O C2O4-2  2CO2 + 2e- 10e- + 16H+1 + 2MnO4-1  2Mn+2 + 8H2O X2 C2O4-2  10CO2 + 10e- X5 16H+1 + 2MnO4-1 + 5CO2O4-2  2Mn+2 + 8H2O + 10CO2 So, we learn how much acid is required and how much water is produced. p. 830 - 3

Another example Complete & balance the following equation using half-reactions: Cr2O7-2 + Cl-1  Cr+3 + Cl2 (in acid sol’n) Cr2O7-2  Cr+3 Cl-1 --> Cl2 14H+1 + Cr2O7-2  2Cr+3 + 7HO2 2Cl-1  Cl2 +14 -2 +6 -2 6e- + 14H+1 + Cr2O7-2  2Cr+3 + 7H2O (x3) 2Cl-1  Cl2 + 2e- 6e- + 14H+1 + Cr2O7-2 + 6Cl-1  2Cr+3 + 7H2O + 3Cl2 + 6e- Finally, check to be sure that atoms & charge are still balanced. p. 830 - 3

One more! Complete & balance this equation using half-reactions: Cu(s) + 4H+1 + 2NO3-1  Cu+2 + 2NO2(g) + 2H2O(l) Cu(s)  Cu+2 4H+1 + 2NO3-1  2NO2(g) + 2H2O(l) Cu(s)  Cu+2 4H+1 + 2NO3-1  2NO2(g) + 2H2O(l) 0 +2 +2 0 Cu(s)  Cu+2 + 2e- 2e- + 4H+1 + 2NO3-1  2NO2(g) + 2H2O(l) 2e- + 4H+1 + 2NO3-1 + Cu(s)  2NO2(g) + 2H2O(l) + Cu+2 + 2e- p. 830 - 3

Last one! Complete & balance this equation using half-reactions: 2Mn+2 + 5NaBiO3 + 14H+1  2MnO4-1 + 5Bi+3 + 5Na+1 + 7H2O(l) 2Mn+2 + 8H2O  2MnO4-1 + 16H+1 5NaBiO3 + 30H+1  5Bi+3 + 5Na+1 + 15H2O +4 +14 +30 +20 2Mn+2 + 8H2O  2MnO4-1 + 16H+1 + 10e- 10e- + 5NaBiO3 + 30H+1  5Bi+3 + 5Na+1 + 15H2O 10e- + 5NaBiO3 + 30H+1 + 2Mn+2 + 8H2O  5Bi+3 + 5Na+1 + 15H2O + 2MnO4-1 + 16H+1 + 10e- 14 7 p. 830 - 3