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UT CH 302 - Study Notes
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(Hint: ln 2 = 0.6931... ~= 0.7)1. Balance the following half reaction in an acid. How many hydrogen ions are needed?Cr(OH)3→ CrO42-a. 5b. 4c. 3d. 2e. 1Cr(OH)3+ H2O --> CrO42-+ 5H++ 3e-2. The reaction below occurs in a(an) _________ cell and the sign of the anode is ______.Cu(s)+ 2Ag+→ CU2++ 2Ag(s)a. Electrochemical, Positiveb. Voltaic, Positivec. Electrochemical, Negatived. Galvanic, NegativeE = 0.8 - (0.2) = 0.6V. The cell is therefore voltaic. The anode loses electrons regardless ofwhether not you have a galvanic or electrochemical cell. It is therefore designated to havenegative sign due to a lose of electrons3. Which statements are true in regards to a table of standard half cell reduction potentials?I. The reactants of the reaction are oxidizing agentsII. The more positive the potential the better the reducing agentIII. The reactions shown are reductionsIV. The oxidizing number of the products is smaller than the reactantsa. I, III, IVb. I, II, III, IVc. II, IIId. II, IVThe reactants in a reduction are oxidizing agents. The reactants in oxidations are reducingagents.4. If a dead battery has a ratio of 2 for [Fe2+]/[Cd2+], what is the standard cell potential?(Hint: log(2) = 0.301)a. 0.36 Vb. 0.33 Vc. -0.33 Vd. -0.36 VEcell=Eocell+ 0.05916/n * log([products]/[reactants])0 = X + 0.05916/2 * log(2)X = -0.331 V5. Rank the following from weakest to strongest oxidizing agent: Li+, K+, Mg+2, Zn+2Li++ e-→ Li -3.1 VK++ e-→ K -2.9 VMg+2+ 2e-→ Mg -2.4 VZn+2+ 2e-→ Zn -0.76 Va. Li+< K+< Mg2+< Zn2+b. Li+< Mg+2< K+< Zn2+c. Zn+2< Mg+2< K+< Li+d. Zn+2< Mg+2< K+< Li+The higher the reduction potential, the stronger the oxidizing agent.6. How many moles of Al3+were needed to produce 4 moles of Al when 0.5 faraday ofcharge passes through a cell?a. 0.2 mol Al3+a. 4 mol mol Al3+b. 1 mol3+c. 0.1 moles Al3+1 faraday corresponds to 1 mol e-. Therefore, .5 faradays is due to .5 mol e-. In order toproduce Al from Al3+3 electrons are needed.x moles Al3+= (.5 mol e-/4 mol Al) = .11 moles Al3+7. What current is needed to produce 98.5 g of solid gold from Au+ in 2 hours?a. 6.7 Ab. 24.1 kAc. 804.0 Ad. 7.0 kAI = (Moles of Product)(Moles of electrons)(Faraday)/timeI = 0.5 moles * 1 e- *96,485 / (7200 sec)I = 6.7 A8. The following reaction is at equilibrium. If [Ag+] = 1x10-6M, what is [Cu2+]?Cu2++ 2Ag(s) → Cu(s)+ 2Ag+a. 5x10-9Mb. 5x10-15Mc. 2x108Md. 5x10-7 MEo= (0.2) - .8 = -0.6VE = Eo- (0.05916/n)*logK0 = Eo- (0.05916/n)*logK-0.6V = (0.05916/n)*log([Ag+]2/[Cu2+])(1x10-6M)2/[Cu2+] = 10^(2*-0.6/0.05916)9. Which of the following is not true about nickel-metal hydride batteries?a. They are not rechargeableb. Nickel is the cathodec. They have 2-3 times more capacity than a NiCd battery of equivalent size.d. A hydrogen absorbing alloy is the anode10. Consider the reaction below:2A + B → ½CIf [C] increases from 0 M to 0.5 M in the course of 1 minute, what [B] remains after 30seconds if it is initially 1 M?a. 0.5 Mb. 0 Mc. 1 Md. 1.5 MFor this reaction, becuse the stoichiometric ratio of B to C is 2:1, B will be used at twice therate C is produced. So, of [C] increases by 0.5 M in 1 minute, [B] will decrease by twice thatamount, 1 M, in 1 minute. After only 30 seconds, it will be half as much.11. Consider the data below:Experiment [A] (M) [B] (M) [C] (M)initial rate(M∙s-1)1 1 0.5 2 0.42 2 3 2 0.83 1 0.5 1 0.84 2 0.5 4 0.4If only these species are pertinent, what is the overall order of the reaction?a. 1stb. 3rdc. 0d. 2ndIn general, ([A]x/[A]y)a∙([B]x/[B]y)b∙([C]x/[C]y)c= (ratex/ratey)The trick to making use of this relationship is to pick sets of experiments (x and y), wherethere is only one unknown value, the order with respect to a single species. For example inexperiments 1 and 3, only [C] is varied, so the expression reduces to:([C]x/[C]y)c= (ratex/ratey)(2/1)c= (0.4/0.8)c = -1Now that we know this, solving other such expressions is easier, because knowing c, theorder with respect to C, allows us to include it in solving other expressions.One can also take the approach of inference. For example, In experiments 1 and 4 thespecies [B] is unchanged and both [A] and [C] are doubled. The rate is unchanged. Havingalready determined that the reaction is -1st order with respect to C, we can infer that thereaction must be 1st order with respect to A, such that the two effects of doubling bothspecies cancel out and the observed initial rate remains unchanged.I'll leave it to you to determine the reaction order with respect to B.12. Consider the following elementary reaction:O(g) + O(g) → O2(g)If [O]ois 4 atm and after 3 seconds [O] has decreased to 1 atm, what is the value of k, therate constant, for this reaction?a. 0.250 M-1∙s-1b. 0.375 M-1∙s-1c. 0.500 M-1∙s-1d. 0.125 M-1∙s-1Because the reaction is an elementary reaction, we can infer that it must be second order,as O must collide with another O to make O2.[O]-1= [O]o-1+ akt1-1= 4-1+ 2k30.75 = 6kk = 0.125 M-1∙s-113. integrated rate law calculation (half life)Consider the following reaction:H2O2(aq) → H2O(l) + ½O2(aq)If, at a certain temperature and in the presence of a catalyst, the rate constant for thisdecomposition is 0.00726 s-1, what is the apporximate half life of hydrogen peroxide?(Note: the actual catalyzed rate of decomposition of hydrogen peroxide is not first order andis in fact rather complicated, but assume it is first order here to make the calculationsdoable.)a. 50 secondsb. 100 secondsc. 200 secondsd. 500 secondse. 1000 secondst½= ln 2/akt½= 0.6931/0.00726 s-114. In a straight-line plot for a 2nd order reaction, the x-axis has units of ______ and the y-intercept is the ______ of an initial concentration.a. concentration, natural logb. concentration, inversec. time, natural logd. time, inverse15. All of the the factors below can increase the rate of a reaction. However, one of them isnot considered in collision theory. Which is it?a. increased reactant concentrationb. addition of a catalystc. increased temperatured. all are considered in collision theoryCollision theory does not speculate about mechanisms and thus does not concern itself withthe impact of catalysts on rates.16 According to transition state theory, a set of chemical species that are involved in anequilbrium will be predcited to spend the least amount of time as ______.a. reactantsb. productsc. transition statesd. transition state theory doesn't allow for such predictionsThe most energetic species is by definition the least stable and thus the least


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