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Chem 222 Exam 3 Study Guide

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by: Leslie Pike

Chem 222 Exam 3 Study Guide Chem 222

Marketplace > Western Kentucky University > Chemistry > Chem 222 > Chem 222 Exam 3 Study Guide
Leslie Pike
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Everything in Chapters 18 and 19: entropy, enthalpy, Gibbs free energy, balancing redox reactions, determining oxidation states, galvanic cells, electrolytic cells, electrolysis
College Chemistry 2
Darwin Dahl
Study Guide
Chemistry, General Chemistry, Redox reactions, Electrochemistry
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This 8 page Study Guide was uploaded by Leslie Pike on Friday April 29, 2016. The Study Guide belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 59 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.


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Date Created: 04/29/16
Chem 222 Study Guide for Exam 3 Chapter 18 Entropy A reaction is spontaneous if it happens of its own accord, without requiring any outside work. Iron rusting is a spontaneous reaction. Sugar dissolving in water is spontaneous. Spontaneous reactions can be exothermic or endothermic, but ALL spontaneous reactions result in a loss of energy. ΔG is negative. If ΔG is positive, that means that you have to add energy to the system to make the reaction proceed in that direction (i.e. the reaction is not spontaneous.) Spontaneous reactions usually have an increase in entropy, symbolized by ΔS. Entropy is the measure of the disorder of a system. When order increases, entropy decreases. When disorder increases, entropy increases. Do not confuse entropy with enthalpy. Enthalpy is the internal energy (or internal heat) of a system. Remember than enthalpy is heat, which you can measure with a thermometer. Enthalpy is symbolized ΔH. The entropy of a solid is less than the entropy of a liquid, which is much less than the entropy of a gas. A gas has very, very high entropy. If the change in entropy is positive, the system is becoming more disordered. If the change in entropy is negative, the system is becoming more ordered. A positive ΔS means that more disorder has been added to the system. A negative ΔS means that disorder has been lost (the system has become more ordered). Entropy can be calculated by: S=k∗ln (W) Here, k is a constant. W is the number of microstates of a system. As W increases, so does entropy. To illustrate: I have two pairs of shoes at school with me. I’m not the most organized person and my shoes tend to get strung out all over the room, but since I only have four shoes total, they don’t make much of a mess. I have four microstates (four shoes) and low entropy. My roommate has lots of shoes—let’s say 7 pairs. She is not the most organized person either, but her shoes make a considerable mess on account of there being so many of them. She has fourteen microstates (fourteen shoes) and high entropy. Basically, the more stuff you have, the bigger potential mess you can make, and the higher the entropy. Thus, bigger molecules will have higher entropies than smaller ones (bigger molecules have more protons, neutrons, and electrons, and thus more microstates). At absolute zero, the entropy is zero. This is because, at absolute zero, there is only one microstate (everything is completely still) and the natural log of one is zero. Adding more heat to a system will cause an increase in entropy (ΔS will be positive) because things move around more when they are heated, meaning you get more microstates. A phase change of a solid to a liquid or gas gives ΔS > 0 because liquids and gases are more disordered than solids, and vice versa. Any element in its elemental state has an enthalpy of formation (ΔH ) of 0.fThus, graphite, O 2 and other standard states have no ΔH and they can be ignored in calculations of ΔH .rxnese elements in their standard states DO have a ΔS, because ΔS and ΔH are not the same thing. ΔH is energy gained or lost (aka heat, thus the capital H), and ΔS is disorder gained or lost. You can remember that disorder is symbolized with a capital S because the word disorder contains the letter “s” and the word “heat” doesn’t. For a given reaction, increase in the number of moles of gas means a positive ΔS. Decrease in the number of moles of gas means a negative ΔS. When no gas is involved, ΔS will be small. (Remember this if we get a test question that asks which reaction has the least ΔS; the correct answer will be the reaction which involves no gas molecules.) 0 ΔS rxns the standard entropy change for a reaction that takes place at 1 atm and 298 K. It can be calculated by the sum of the ΔS of the products, minus the sum of the ΔS of the reactants. Remember to multiply by the balancing coefficients. Gibbs Free Energy To determine whether or not a reaction is spontaneous at a given temperature, we calculate the Gibbs free energy for that reaction. Gibbs free energy is calculated with the following equation: ∆ G=∆H−T ∆S If the ΔG is negative, that means that energy is released by the reaction when it proceeds, and thus the reaction is spontaneous. Imagine a child on a sliding board. The child slides down spontaneously (without any outside work). The child loses energy in the act of sliding down (potential energy of the child at the top is transferred to the slide and the child’s shirt as heat). If ΔG is positive, that means that energy has to be added in order for the reaction to proceed. The child is not going to slide up the sliding board unless somebody is pushing him or her. If ΔG is zero, the reaction is at equilibrium and it does not proceed either way. A child sitting at the bottom of a sliding board won’t go up or down. 0 ΔG fis the standard free energy of formation of one mole of a material. It is calculated by summing the ΔG’s of one mole of all the elements that make up the material. Superscript zeros mean standard conditions, temperature 298 K and pressure 1 atm. Sample problem: Calculate ΔG for a given reaction. ΔS is 42 J/(K*mol) and ΔH is 0 35 kJ/mol. First step: figure out what you have, what you need, and what equation you can use to get from what you have to what you need. You have ΔS and ΔH, and you need ΔG. So, you can use the following equation: ∆G=∆H-T∆S This looks fine, except you weren’t given T. That’s okay, you don’t need to be given T. The superscript zeroes should clue you in to the fact that you are at standard conditions, which means that T is 298 K. So, you can just start plugging in numbers now, right? WRONG!!!! Before you plug anything in, make sure that it is in the proper units. ∆S was given to you with J, and ∆H with kJ. Convert kJ to J before plugging in your numbers. ∆G=∆H-T∆S=35000-298*42=22484 J/mol The equilibrium constant can be calculated from the Gibbs free energy with the following equation: ∆ G =−RT∗ln(K) Here, R=8.314 J/(mol*K). Common pitfall: be sure to convert C to K and kJ to J before you start plugging numbers in. 0 If K is greater than 1, the natural log will be positive, ∆G will be negative, and the reaction proceeds spontaneously to the right. If K is less than 1, the natural log will be negative, ∆G will be positive, and the reaction will not proceed spontaneously to the right (in fact, it will proceed spontaneously to the left). Chapter 19 Electrochemistry deals with redox reactions. Electrochemistry deals with one of two situations: a reaction making electricity, or electricity being used to force a reaction to go in a certain direction. Oxidation states Redox reactions involve a change in oxidation states. An oxidation state is the electrical charge on an atom or molecule.  Free elements (Pd ,(s) 2 (g)have an oxidation state of 0 because they have no charge.  Monatomic ions have an oxidation state equal to their superscript number (Fe3+ has an oxidation state of +3)  Alkali metals (in solution or in compounds) have an oxidation state of +1  Alkali earth metals (in solution or in compounds) have an oxidation state of +2  Fluorine (unless it is a gas) ALWAYS, ALWAYS, ALWAYS has an oxidation state of -1. All other halogens also have an oxidation state of -1, unless they are bonded to fluorine.  Oxygen (unless it is a gas) has an oxidation state of -2. As peroxide, the oxygens have a charge of -1 apiece (for a total charge of -2).  Hydrogen has an oxidation state of +1 when it is written at the beginning of the compound, and -1 when it is written at the end of the compound.  All other elements have whatever charge it takes to balance out the charges of the above atoms. For instance, in K CrO , t2e p4tassium contributes a charge of +2 and the oxygen contributes a charge of -8. Chromium therefore must have a charge of +6 to make the molecule neutral. Oxidation states do not have to be whole numbers. Redox reaction balancing Balancing a redox reaction (acidic aqueous solution): Cr2O 72-+ Fe 2+ Cr 3++ Fe 3+ Split it into two half-reactions. 2- 3+ Cr2O 7  Cr Fe2+  Fe3+ You will have to balance both equations. Start with the easier reaction (in this case, the one with iron). First, mass balance. (Already done.) Then, charge balance. Charge balancing looks intimidating when you first try it, but actually it’s just simple arithmetic. The left side and the right side have to equal. You have +2 on the left and +3 on the right. To make the equation balanced, subtract 1 from 3. +2 = +3 -1 Ta-da, it’s that easy. You add one electron to the right side of the equation, and you’re done! Now, for the other half-reaction. Mass balance it first. (You are in aqueous solution so you can add all of the water molecules you need, since you have plenty of them floating around. You are in acidic solution so you can add all of the hydrogen ions you need, since you have plenty of them floating around.) + 2- 3+ 14H + Cr O 2 7  2Cr + 7H O2 Now, you need to charge balance it. This equation looks more intimidating than the previous one, but you can just break it down into smaller pieces. Hydrogen, oxygen, and water do not participate in the actual electron transfer—they can be ignored. This leaves us with chromium. The chromium in dichromate has a charge of +6, and the free chromium ion has a charge of +3. DO NOT NEGLECT THE FACT THAT THERE ARE TWO CHROMIUM ATOMS IN DICHROMATE, AND TWO FREE CHROMIUM IONS ON THE RIGHT SIDE OF THE EQUATION. Write the amount of charge on each side of the equation, and subtract where you need to subtract to make both sides equal (electrons) + 2(+6) = 2(+3)  -6 +12 = +6 You have to add six electrons to the left. Check your work; one of the half-reactions should have electrons on the right, and the other should have them on the left. (This is why you do the easy one first, so you can check the harder one against the easy one.) The electrons will need to cancel. You have 6 electrons in the chromium equation, and 1 electron in the iron equation. Multiply the iron equation by 6 so that the electrons cancel, and add the two equations. + 2- 2+ 3+ 3+ 14H + Cr O 2 7 + 6 Fe  2Cr + 7H O2+ 6Fe This is how to balance a redox reaction in acidic aqueous solution. For a basic aqueous solution, do the same thing. Then, after you have the final equation, add + hydroxide to both sides to cancel the H . For the above equation, this would be 14 hydroxide. You would have 14 water on the left, 7 water on the right, and 14 hydroxide on the right. Waters cancel each other, and you have 7 water on the left and 14 hydroxide on the right. Electrochemical cells There are two types of electrochemical cells: galvanic (voltaic) or electrolytic. In a galvanic cell, the reaction makes electricity. In an electrolytic cell, electricity makes the reaction. Galvanic cells have a positive EMF, and electrolytic cells have a negative EMF. Oxidation occurs at the anode, and reduction at the cathode. (The word “oxidation” and the word “anode” both start with vowels, and the word “reduction” and the word “cathode” both start with consonants.) Anodes shrink and cathodes grow. Oxidation is the loss of electrons (metal turns to aqueous ions), and reduction is the gain of electrons (aqueous ions turn to solid metal). When writing the shorthand notation for a battery, you start at the anode and end at the cathode. (A comes before C in the alphabet, so you do the anode first.) For the traditional zinc-copper battery: 2+ 2+ Zn (s) (aq)(x M)||Cu (aq)y M)|Cu (s) Here, x and y represent the concentrations of the zinc and copper ions, respectively. The standard EMF of the cell is calculated as: 0 0 0 E cell E cathode (-E anode The cathode reaction will be going in the forward direction as it is written in a table. The anode reaction will be going in the reverse direction as it is written in the table; hence, its EMF will have an opposite sign of whatever the table says (this is why it has a negative sign in front of it in the equation). Standard conditions are 298 K, 1 M concentration and 1 atm. To determine the EMF of a cell not at standard conditions: 0 E = E – (.0592/n)*log(Q) 0 0 Calculate Q for your reaction, calculate E cellrom the E values of your cathode and anode, and determine the number of free electrons involved in the reaction. Then, plug everything in and solve. (I personally do not recommend solving for E cathodend Eanodeseparately; this creates extra steps and extra potential for mistakes.) To convert from free energy to EMF: ΔG=-nFE F is the Faraday constant, 96,485 Coulombs/mole of electrons. Lowercase n is the number of free electrons involved in the reaction, E is cell EMF. Sample problem: Calculate ΔG for the following reaction at standard conditions: + 2+ 2Ag (aq)+ Cu (s)> 2Ag (s) Cu (aq) The standard reduction potential for silver is 0.799, and the standard reduction potential for copper is 0.337. First step: figure out what you have, what you need, and what relationship gets you 0 from what you have to what you need. You have the E for your two reactions, and you need ΔG. ΔG is related to E by the following equation: ΔG=-nFE. E for the complete reaction is the sum of the E of the two half-reactions. One half reaction is going forward, and the other is going backward. In the equation at the top, silver is written as a reduction; thus, its reduction potential is the same as given in the table. DO NOT multiply by the balancing coefficient. Copper is being oxidized in the above reaction, not reduced, meaning that its E will have an opposite sign of the number given in the table. Add the two half-reactions to get the E for the full reaction: Ecell +0.799 + (-0.337) = 0.462V. Plug this value into the expression for ΔG (n is 2 because you have 2 electrons being transferred in the above reaction, and F is Faraday’s constant, 96485): ΔG=-nFE = -(2)(96485)(.462)= -89000 J In an electrolytic cell, an electric current drives a nonspontaneous reaction. Electrolytic cells have a negative E, this is how you distinguish an electrolytic cell from a galvanic cell. Positive E means spontaneous, negative E means nonspontaneous. (Notice that this is the opposite of what the sign means when you are talking about free energy.) In an electrolytic cell, electrons are added to the cathode and removed from the anode. Metal plates out on the cathode, because it is reduced from its ionic form to its solid form when it takes the extra electrodes from the cathode. Whatever anion the metal was bounded to donates its electrons to the anode. Sample problem: You are purifying copper by electrolysis. You need 2 kg of pure copper. You have a 10 amp current source you can plug your cathode and anode into, and an unlimited supply of impure copper. How long does it take for 2 kg of pure copper to plate out onto the cathode? This problem looks hairy, but trust me, it’s nothing you haven’t already done in Chem 120. This is a regular stoichiometry problem; you just need to remember that a copper ion is missing two electrons (two moles of electrons are needed per one mole of copper), and you need to learn two new relationships:  96,485 C =1 mole electrons  An amp is a coulomb per second (C/s), making a 10 amp current 10 C/ 1 s. 2000 g copper * (mol copper/63.55 g copper) * (2 mol electrons/mol copper) * (96485 coulombs/mol electrons) * (second/10 coulombs) = 607301 seconds = 168 hours = 7 days. You will be waiting a whole week for that copper. Better find a stronger current source. Common pitfalls to avoid  Don’t forget to raise to the balancing coefficients when calculating Q  Convert kilojoules to joules before plugging numbers into equations  Convert Celsius to Kelvin before plugging numbers into equations  Half-reactions are written as reductions, and their reduction potentials are given. Reverse the sign for the reaction that is an oxidation in the main reaction, but don’t reverse it twice! (i.e. don’t multiply by -1 and then subtract, do one or the other)  Don’t multiply reduction potential by the balancing coefficient (or raise it to the balancing coefficient) Important things to memorize  96,485 C =1 mole electrons (this is equivalent to Faraday’s constant, F)  An amp is a coulomb per second (C/s), making a 10 amp current 10 C/ 1 s.  ∆G=∆H-T∆S  ΔG= --nFE 0 0 0  E cell E cathode (-E anode  E = E – (.0592/n)*log(Q) 0  ∆G = --RT*ln(K)


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