18 2 18 3 Electrochemistry Balancing Redox Equations

18. 2 -18. 3 Electrochemistry. Balancing Redox Equations By: Maisy Holehouse

Balancing Oxidation-Reduction Equations ● Oxidation State- the charge of an an atom that has lost or gained electrons ● Oxidation - loss of electrons, also can mean an increase in oxidation state ● Reduction - gain of electrons, also can mean a decrease in oxidation state ○ Leo the Lion goes Ger (loss electrons oxidation, gain electrons reduction)

Continued. . . Remember: Mass and charge must be conserved in a chemical equation Half- reaction method of Balancing: To balance an equation, first we must separate the original reaction into two separate equations, the oxidation reaction and the reduction reaction. There are slightly different steps if you are balancing in an acidic or basic solution

Assigning Oxidation Numbers: Please look at the sheet that is being passed out

Steps to Balance an equation 1. 2. 3. 4. Assign Oxidation States Separate the entire reaction into two half reactions Balance each of the half reactions by mass Add in O to make the reaction equal, then add in enough H 20’s for each oxygen, then add the corresponding H+’s on the opposite side. (For basic solutions: after adding the H+’s, add enough OH - to neutralize the effect of the H+) 5. Balance each of the half reactions by charge 6. Make the number of electrons in both half reactions equal 7. Add two half reactions together

Balancing Redox Reactions in Acidic Solutions Fe 2+ (aq) + Mn. O 4 - (aq) -> Fe 3+ (aq) + Mn 2+ (aq) +2 +2 +7 -2 Reduction +3 oxidation Oxidation: Fe 2+ (aq) -> Fe 3+ (aq) + 1 e. Reduction: Mn. O 4 - (aq) -> Mn 2+ (aq) + 4 H 2 O (l) 8 H+ (aq) + Mn. O 4 - (aq) -> Mn 2+ (aq) + 4 H 2 O (l) 5 e- + 8 H+ (aq) + Mn. O 4 - (aq) -> Mn 2+ (aq) + 4 H 2 O (l)

Continued … 5 [Fe 2+ (aq) -> Fe 3+ (aq) + 1 e-] 5 Fe 2+ (aq) -> 5 Fe 3+ (aq) + 5 e. Combine: 5 Fe 2+ (aq) -> 5 Fe 3+ (aq) + 5 e- + 8 H+ (aq) + Mn. O 4 - (aq) -> Mn 2+ (aq) + 4 H 2 O (l) Final: 5 Fe 2+ (aq) + 8 H (aq) + Mn. O 4 - (aq) -> 5 Fe 3+ (aq) + Mn 2+ (aq) + 4 H 2 O (l)

Balancing Redox Reactions in Basic Solutions 2 I- (aq) + Mn. O 4 - (aq) -> I 2 (aq) + Mn. O 2 (s) -1 +7 -2 0 Reduction +4 -2 Oxidation: 2 I- (aq) -> I 2 (aq) Reduction: Mn. O 4 - (aq) -> Mn. O 2 (s) + 2 H 2 O 4 H+ (aq) + 4 OH- (aq) + Mn. O 4 - (aq) -> Mn. O 2 (s) + 2 H 2 O (l) + 4 OH- (aq) H 2 O (l)

Continued … Balance by charge: 2 I- (aq) -> I 2 (aq) + 2 e 4 H 2 O (l) + Mn. O 4 - (aq) + 3 e- -> Mn. O 2 (s) + 2 H 2 O (l) + 4 OH- (aq) Make electron numbers equal: 3 [2 I- (aq) -> I 2 (aq) + 2 e-] = 6 I- (aq) -> I 2 (aq) + 6 e 2 [4 H 2 O (l) + Mn. O 4 - (aq) + 3 e- -> Mn. O 2 (s) + 2 H 2 O (l) + 4 OH- (aq)] = 8 H 2 O (l) + 2 Mn. O 4 - (aq) + 6 e- -> 2 Mn. O 2 (s) + 4 H 2 O (l) + 8 OH- (aq) Add together: 6 I- (aq) -> I 2 (aq) + 6 e 4 8 H 2 O (l) + 2 Mn. O 4 - (aq) + 6 e- -> 2 Mn. O 2 (s) + 4 H 2 O (l) + 8 OH- (aq) 6 I- (aq) + 4 H 2 O (l) + 2 Mn. O 4 - (aq) -> 2 Mn. O 2 (s) + I 2 (aq) + 8 OH- (aq)

Sample Problem Balance the reaction in an Acidic solution: Al (s) + Cu 2+ (aq) -> Al 3+ (aq) + Cu (s) Answer: 2 Al (s) + 3 Cu 2+ (aq) -> 2 Al 3+ (aq) + 3 Cu (s)

Voltaic Cell:

Voltaic Cells: Generating Electricity from Spontaneous Chemical Reactions ● ● ● ● Electrical Current: Flow of electric charge Electrochemical cell: The way electricity is generated through redox reactions Voltaic Cell: an electrochemical cell that produces electrical current from a spontaneous reaction Electrolytic cell: an electrochemical cell that consumes energy to make a nonspontaneous chemical reaction occur. Half-Cell: one half, or one jar of the voltaic cell Electrodes: metal strips placed into the half cells that are conductive surfaces, to let electrons flow through Potential Difference: A measure in the difference of potential energy per unit of charge Electromotive Force: Potential difference/ motion of electrons.

● ● ● In a Voltaic Cell the potential difference between the two electrodes is the cell potential. (Ecell) or cell emf, the more willing a substance is to be either oxidized or reduced, the higher the cell potential. (found in appendix 2) ○ Cell potential: Measure of ability of redox reaction to happen spontaneously, negative means the forward reaction is not possible. Voltaic cells are dependent on temperature and concentration (Standards: 25 C, 1 M solutions, 1 atm gas) This is called the standard cell potential (E*cell) Electrode with oxidation: anode (-) ; Electrode with reduction: cathode (+) ○ Electrons flow anode to cathode, negative to positive (red cat/ an ox) Cations flow out of the anode (because electrons are flowing out, they are not there to neutralize the ions), there is a build up of positive charge in the solution, and vise versa for the cathode, electrons flow into it, there is a buildup of negative charge. To allow the reaction to continue, there is a salt bride suspended, one end into each half cell. This salt bridge has ions, the cations go to the half cell with the cathode, or the greater amount of negative charge; the anions flow to the cell with the anode, to neutralize the build up of positive charge.

Zinc and Copper Questions A galvanic reaction is going to take place between zinc and copper. The Ecell=1. 10 V Can anyone tell me, If Zinc is more willing to give up electrons (more easily oxidized) ● ● ● ● Which one is being oxidized? Reduced? (and which one is the anion/ cation) Zn: oxidized/anion, Cu: reduced/cation The half reactions for each (next slide) The equation, with charges Zn (s) + Cu 2+ (aq) -> Cu (s) + Zn 2+ The directions of the flow of the electrons (From which element to which element)

Math of Electrochemistry Amperes (A/ amps): the unit in which electrical current is measured, one amp represents the flow of one coulomb (C) per second. 1 A = 1 C/s An electron has a charge of 1. 602 x 10 -19 C. 1 A = 6. 242 x 1018/s this flow of electrons is caused by Potential difference, which is the potential energy difference measurement, and the unit for this is Volts (V) 1 V = 1 J/C This means the difference of one volt shows the charge of a coulomb has an energy difference of one joule between two electrodes.

Electrochemical cell Notation ● ● Cell Diagram/ Line notation: ○ Oxidation on the left, and reduction on the right, with a double vertical line in between (salt bridge and indication the separation) ○ Ex: Zn 2+/Cu 2+-> Zn(s) Zn 2+ (aq) Cu (s) ○ Compounds/elements in different phases are separated by a vertical line (a boundary between phases) ○ In some, the reactants/ products are in the same phase, so we use a comma to separate them Inert Electrode: Used as an anode or cathode, and helps with electron transfer but it not part of the actual reaction (Pt is commonly used)

Sample Problems: Balance in a basic Solution: Al (s) + Mn. O 4 - (aq) -> Mn. O 2 (s) + Al(OH)4 - (aq) Answer: Al (s) + Mn. O 4 - (aq) + 2 H 2 O -> Mn. O 2 (s) + Al(OH)4 - (aq)

Sample Problem: Draw a voltaic cell for each reaction, Label anode and cathode, show the half-reaction at each electrode. And the direction flow: 2 Ag+ (aq) +Pb (s) -> 2 Ag (s) + Pb 2+ (s) Answer:

College Board: