CHEM 180/181Chapter 20 Dana Roberts Chapters covered: 18 and 20 Notes available online or in Resource Room (1 st floor). To print notes: From print window, select “print handouts” and 4 per page in black and white. Should turn out fine. Need additional help? me to set up an appointment. I’m almost always available!

CHEM 180/181Chapter 20 Chapter 20 Electrochemistry

CHEM 180/181Chapter 20 Topics in Electrochemistry Oxidation-Reduction (Redox) Numbers Balancing Redox Reactions Voltaic Cells Cell EMF Spontaneity of Redox Reactions Batteries Corrosion Electrolysis

CHEM 180/181Chapter 20 Oxidation and Reduction Oxidation (loss of e - ) Na Na + + e - Reduction (gain of e - ) Cl + e - Cl - Oxidation-reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. Electron transfer can produce electrical energy spontaneously, but sometimes electrical energy is used to make them occur (nonspontaneous).

CHEM 180/181Chapter 20

CHEM 180/181Chapter 20 Oxidation-Reduction (Redox) Reactions BOTH reduction and oxidation must occur. A substance that gives up electrons is oxidized and is called a reducing agent or reductant (causes another substance to be reduced). A substance that accepts electrons is reduced and is therefore called an oxidizing agent or oxidant (causes another substance to be oxidized).

CHEM 180/181Chapter 20 Determine if it’s a redox reaction by keeping track of the oxidation states of all elements involved. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) (0) (+1) (+2) (0) Quick hint: If a reaction includes an ELEMENT that turns into an ion, it’s a redox reaction. Oxidation-reduction (Redox) Reactions

CHEM 180/181Chapter 20 1.Atoms in elemental form, oxidation number is zero. (Cl 2, H 2, P 4, Ne are all zero) 2.Monoatomic ion, the oxidation number is the charge on the ion. (Na + : +1; Al 3+ : +3; Cl - : -1) 3.O is usually -2. But in peroxides (like H 2 O 2 and Na 2 O 2 ) it has an oxidation number of H is +1 when bonded to nonmetals and -1 when bonded to metals. (+1 in H 2 O, NH 3 and CH 4 ; -1 in NaH, CaH 2 and AlH 3 ) 5.The oxidation number of F is The sum of the oxidation numbers for the molecule is the charge on the molecule (zero for a neutral molecule). Oxidation Number Guidelines

CHEM 180/181Chapter 20 Example Determine the oxidation state of all elements in ammonium thiosulfate (NH 4 ) 2 (S 2 O 3 ). (NH 4 ) 2 (S 2 O 3 ) NH 4 + S 2 O total +2 NH 4 + S 2 O sum = overall charge on ion -3 +4

CHEM 180/181Chapter 20 Determining Oxidation States What is the oxidation state of Mn in MnO 4 - ? Answer: +7

CHEM 180/181Chapter 20 Balancing oxidation-reduction equations Balancing chemical equations follows law of conservation of mass. AND, gains and losses of electrons must also be balanced.

CHEM 180/181Chapter 20 Half-Reactions Separate oxidation and reduction processes in equation, Sn 2+ (aq) + 2Fe 3+ (aq) Sn 4+ (aq) + 2Fe 2+ (aq) Oxidation: Sn 2+ (aq) Sn 4+ (aq) + 2e - Reduction: 2Fe 3+ (aq) + 2e - 2Fe 2+ (aq) Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half reaction.

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic Consider : MnO 4 - (aq) + C 2 O 4 2- (aq) Mn 2+ (aq) + CO 2 (g) Unbalanced half-reactions: MnO 4 - (aq) Mn 2+ (aq) C 2 O 4 2- (aq) CO 2 (g) First, balance everything EXCEPT hydrogen and oxygen. Deal with half-reactions SEPARATELY. (acidic)

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic To balance O: Add 4H 2 O to products to balance oxygen in reactants. MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) To balance H: Add 8H + to reactant side to balance the 8H in water. 8H + (aq) + MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l)

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic Balance charge: Add up charges on both sides. Add 5 electrons to reactant side. 5e - + 8H + (aq) + MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) Mass balance of C in oxalate half-reaction. C 2 O 4 2- (aq) 2CO 2 (g) Balance charge by adding two electrons to the products. C 2 O 4 2- (aq) 2CO 2 (g) + 2e - Last step: Cancel electrons and add reactions together.

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic 5e - + 8H + (aq) + MnO 4 - (aq) Mn 2+ (aq) + 4H 2 O(l) C 2 O 4 2- (aq) 2CO 2 (g) + 2e - Top reaction times 2. Bottom reaction times 5. ALL PARTS!! 10e H + (aq) + 2MnO 4 - (aq) 2Mn 2+ (aq) + 8H 2 O(l) 5C 2 O 4 2- (aq) 10CO 2 (g) + 10e -. 16H + (aq) + 2MnO 4 - (aq) + 5C 2 O 4 2- (aq) 2Mn 2+ (aq) + 8H 2 O(l) + 10CO 2 (g) **Electrons have cancelled out**

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Acidic Summary 1. Divide equation into two incomplete half-reactions. 2. Balance each half-reaction (a) balance elements other than H and O. (b) balance O atoms by adding H 2 O. (c) balance H atoms by adding H+ (basic conditions will require further work at this step). (d) balance charge by adding e - to the side with greater overall positive charge.

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Summary 3.Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other half-reaction. 4. Add the two half-reactions and cancel out all species appearing on both sides of the equation. 5. Check equation to make sure there are same number of atoms of each kind and the same total charge on both sides. Errors can be caught!!

CHEM 180/181Chapter 20 Balancing Equations by the Method of Half-Reactions: Basic Balancing process is started using H + and H 2 O, then adjusting with OH - to uphold reaction conditions (H + does not exist in basic solutions). Balance the following reaction: H 2 O 2 (aq) + ClO 2 (aq) ClO 2 - (aq) + O 2 (g) (basic)

CHEM 180/181Chapter 20 Basic Redox Reactions Split into two half-reactions. H 2 O 2 (aq) O 2 (g) ClO 2 (aq) ClO 2 - (aq) Balance elements, then oxygen by adding H 2 O. Then, add H + to balance H, just like an acidic redox reaction. H 2 O 2 (aq) O 2 (g) + 2H + ClO 2 (aq) ClO 2 - (aq) Balancing Equations by the Method of Half-Reactions: Basic

CHEM 180/181Chapter 20 Basic Redox Reactions Add OH - to both sides, enough to neutralize all H + (basic reactions cannot support H + ). 2OH - + H 2 O 2 (aq) O 2 (g) + 2H + + 2OH - ClO 2 (aq) ClO 2 - (aq) Combine H + and OH - to form H 2 O. 2OH - + H 2 O 2 (aq) O 2 (g) + 2H 2 O Balancing Equations by the Method of Half-Reactions: Basic

CHEM 180/181Chapter 20 Basic Redox Reactions Balance charge by adding e -. 2OH - + H 2 O 2 (aq) O 2 (g) + 2H 2 O + 2e - 1e - + ClO 2 (aq) ClO 2 - (aq) Multiply each reaction so both have same e -. Then add them together, cancelling where possible. 2OH - + H 2 O 2 (aq) + 2ClO 2 (aq) O 2 (g) + 2H 2 O + 2ClO 2 - (aq) Double-check your answer! Balancing Equations by the Method of Half-Reactions: Basic

CHEM 180/181Chapter 20 Balance this redox equation Cu(s) + NO 3 - (aq) Cu 2+ (aq) + NO 2 (g) (acidic) Ans: Cu(s) + 2NO 3 - (aq) + 4H + (aq) Cu 2+ (aq) + 2NO 2 (aq) + 2H 2 O(l) Solution posted on notice board.

CHEM 180/181Chapter 20 Balance this redox equation NO 2 - (aq) + Al(s) NH 3 (aq) + Al(OH) 4 - (aq) (basic) 2Al(s) + NO 2 - (aq) +OH - (aq) + 5H 2 O(aq) 2Al(OH) 4 - (aq) + NH 3 (aq) Solution posted on notice board.