ELECTROCHEMISTRY RULES FOR ASSIGNING OXIDATION NUMBERS 1 The

ELECTROCHEMISTRY

RULES FOR ASSIGNING OXIDATION NUMBERS 1. The oxidation # of a monatomic ion is equal in magnitude and sign to its ionic charge. Ex. Br 1 = 1; Fe 3+ = +3 2. The oxidation # of hydrogen in a compound is +1, except in metal hydrides, such as Na. H, where it is 1. 3. The oxidation number of oxygen in a compound is 2, except in peroxides, such as H 2 O 2, where it is 1. 4. The oxidation number of an uncombined atom (element) is 0. 5. For any neutral compound, the sum of the oxidation numbers of the atoms must equal to 0. 6. For any polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion

TERMS • Oxidation = losing electrons by a substance • Reduction = gaining electrons by a substance

• Oxidizing agent = a substance that allows something to undergo oxidation. If the oxidizing agent allows a substance to lose electrons, then the oxidizing agent must accept the electrons oxidizing agent is reduced. • Reducing agent = a substance that allows something to undergo reduction. If the reducing agent allows a substance to gain electrons, then the reducing agent must donate the electrons reducing agent is oxidized

BALANCING REDOX EQUATIONS

1. 2. 3. 4. 5. 6. Break into half reactions Work with only one half reaction at a time. Balance for mass. Balance for charge. Repeat for other half reaction. Check number of electrons. – If number of electrons are the same, add the reactions together. – If different, multiple half reactions by number so that the number of electrons are the same. 7. Add reactions together to get the overall reaction. 8. If done in acid or basic conditions, use the following chart if you are “missing” oxygen or hydrogen

GALVANIC CELL

Galvanic cell – a device where chemical energy is changed to electrical energy. – Works with a spontaneous reaction…a reaction that, given enough time, will take place. – Constructed of two cells • One for the oxidation • One for the reduction

– Parts • Two cells, each with solution in them • Two electrodes, which are metal “rods” inserted into the solution. – Anode • the metal that undergoes oxidation • since electrons are produced here, has a negative charge – Cathode • the metal that undergoes reduction • has a positive charge • External wire – Allows electrons to flow from the anode to the cathode • Salt bridge – Contains a salt that will not form a precipitate with either the ions in each cell. – Allows the charge to balance out • its ions travel opposite to the movement of the electrons

Memorization…You must KNOW!!! • Anode is where Oxidation takes place. • Cathode is where Reduction takes place.

• Electromotive force (EMF) = the “pulling” or “driving” force on the electron to move them through the cell. • Unit of electrical potential is the VOLT 1 volt = 1 joule = J 1 coulomb C • A coulomb is fundamental unit for charge.

STANDARD REDUCTION POTENTIAL (SRP)

• To calculate the potential of each half reaction, an arbitrary standard was selected. • Every standard reduction potential is based on the Standard Hydrogen Electrode (S. H. E. ). – Voltaic cells are connected and the flow of the current is noted. – Depending on the direction of flow, the placement of the half cell can be determined based on the S. H. E. – The voltage for the half cell is determined by arbitrary setting the S. H. E. to have a voltage of zero. • Measured at standard conditions. – 1 atm pressure (standard pressure) – 25 OC (298 K) (room temperature) – 1 M solutions • Cell potential has the symbol of E°

AP Question 1971 Quantitative chemical data are often based on arbitrary standards. Discuss this statement with the following data for fluorine (a)The atomic weight of fluorine is 19. 00 (b)Eo, the standard electrode potential, is +2. 65 volts for the half reaction: F 2 + 2 e 2 F

To find the cell potential, sum up the half cell potentials. • If ½ reaction is written for the reverse reaction, change signs on the cell potential. • NO CHANGE IS NEEDED FOR STOICHIOMETRY COEFFICENTS! – Since the volt is energy/charge, increasing the coefficients increases the energy and the charge by the same amount and therefore the voltage is the same. – Energy and charge are extensive properties. – Voltage is an intensive property.

The setup of the Standard Reduction Potential Table Oxidized form + electron reduced form The more positive the value, the better the oxidizing ability of the species.

RELATIONSHIP BETWEEN DG AND E°

F = faraday constant – The charge carried by one mole of electrons – 96485 coulombs per mole of electrons – can use 96500 C/mol e ∆G = n F E° • ∆G = Gibb’s Free Energy • n = number of moles of electrons • F = faraday constant • E° = cell potential Since E° is the value for the spontaneous reaction, the ∆G must be negative and since cell potential are calculated to be positive the negative sing in the equation is there to “make the signs work out!”

NERNST EQUATION

• Used when cells are NOT @standard conditions. Knowing that the reaction quotient is Q and that it is related to the equilibrium constant, Keq, and since: • ∆G = RTln. K and ∆G = n FE��˚ one can produce the following equation: • Using the values of R and T and F and the conversion between ln and log base 10, we have • If the cell is @ equilibrium then Eo = 0 and Q must be equal to Keq, then log 10 K = n. E˚ 0. 0592

Electrical Energy Since electrochemistry deals with moving electrons and current is the amount of charge (coulombs) passed per unit of time (seconds). Current = charge = coulombs = Amps time seconds Faradays constant (F)= charge carried by one mole of electrons = 96485. 3 columbs/mole e We use the balance equation, F, and current to find the amount of material deposited at either electrode.

These type of problems are nothing more than unit conversion problems. Problem: A car battery is listed as 475 “cold cranking amps”, which means that is can produce 475 amps for 30 seconds. If the half reaction is: Pb(s) + SO 42 Pb. SO 4 + 2 e How much lead is consumed in the 30 seconds?

ELECTROLYSIS

• A galvanic cell produces current when an oxidation reduction reaction proceeds spontaneously. This produces electricity from a chemical reaction. • An electrolytic cell does the opposite. It produces a chemical reaction from electricity. • This involves supplying electricity to the cells. • The cell potential in an electrolytic cell will be NEGATIVE. • What ever the cell potential value is for the spontaneous reaction is, sets the lower limit on the amount of energy that you need to apply to the electrolytic cell.

• The reaction between Zn|Zn 2+||Cu 2+|Cu has an E° value of 1. 10 volts. To reverse this reaction, you must have greater than 1. 10 volts applied to do it. To solve the stoichiometry problems you must: 1. Use current and time to find amps 2. Use amps and Faraday’s constant to find moles of electrons. 3. Use the balance equation to find the moles of metal. 4. Convert moles to grams.

1974 B • A steady current of 1. 00 ampere is passed through an electrolytic cell containing a 1 molar solution of Ag. NO 3 and having a silver anode and a platinum cathode until 1. 54 grams of silver is deposited. (a) How long does the current flow to obtain this de posit? (b) What weight of chromium would be deposited in a second cell containing 1–molar chromium(III) nitrate and having a chromium anode and a platinum cathode by the same current in the same time as was used in the silver cell? (c) If both electrodes were platinum in this second cell, what volume of O 2 gas measured at standard tempera ture and pressure would be released at the anode while the chromium is being deposited at the cathode? The current and the time are the same as in (b)

1972 Br 2 + 2 Fe 2+(aq) 2 Br (aq) + 2 Fe 3+(aq) For the reaction above, the following data are available: 2 Br (aq) Br 2(l) + 2 e Eo = 1. 07 volts Fe 2+(aq) Fe 3+(aq) + e Eo = 0. 77 volts __________________________ So, cal/mole o. C Br 2(l) 58. 6 Fe 2+(aq) 27. 1 Br (aq) 19. 6 Fe 3+(aq) 70. 1 __________________________ (a) Determine ∆Srxn (b) Determine ∆Grxn (c) Determine ∆Hrxn