e3ps10key - (Hint: ln 2 = 0.6931. ~= 0.7) 1. Balance the...

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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 CrO 4 2- a. 5 b. 4 c. 3 d. 2 e. 1 Cr(OH) 3 + H 2 O --> CrO 4 2- + 5H + + 3e - 2. The reaction below occurs in a(an) _________ cell and the sign of the anode is ______. Cu (s) + 2Ag + CU 2+ + 2Ag (s) a. Electrochemical, Positive b. Voltaic, Positive c. Electrochemical, Negative d. Galvanic, Negative E = 0.8 - (0.2) = 0.6V. The cell is therefore voltaic. The anode loses electrons regardless of whether not you have a galvanic or electrochemical cell. It is therefore designated to have negative sign due to a lose of electrons 3. Which statements are true in regards to a table of standard half cell reduction potentials? I. The reactants of the reaction are oxidizing agents II. The more positive the potential the better the reducing agent III. The reactions shown are reductions IV. The oxidizing number of the products is smaller than the reactants a. I, III, IV b. I, II, III, IV c. II, III d. II, IV The reactants in a reduction are oxidizing agents. The reactants in oxidations are reducing agents. 4. If a dead battery has a ratio of 2 for [Fe 2+ ]/[Cd 2+ ], what is the standard cell potential? (Hint: log(2) = 0.301) a. 0.36 V b. 0.33 V c. -0.33 V d. -0.36 V E cell =E o cell + 0.05916/n * log([products]/[reactants]) 0 = X + 0.05916/2 * log(2) X = -0.331 V 5. Rank the following from weakest to strongest oxidizing agent: Li + , K + , Mg +2 , Zn +2 Li + + e - Li -3.1 V K + + e - K -2.9 V Mg +2 + 2e - Mg -2.4 V Zn +2 + 2e - Zn -0.76 V a. Li + < K + < Mg 2+ < Zn 2+ b. Li + < Mg +2 < K + < Zn 2+ c. Zn +2 < Mg +2 < K + < Li +
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d. Zn +2 < Mg +2 < K + < Li + The higher the reduction potential, the stronger the oxidizing agent. 6. How many moles of Al 3+ were needed to produce 4 moles of Al when 0.5 faraday of charge passes through a cell? a. 0.2 mol Al 3+ a. 4 mol mol Al 3+ b. 1 mol 3+ c. 0.1 moles Al 3+ 1 faraday corresponds to 1 mol e-. Therefore, .5 faradays is due to .5 mol e-. In order to produce Al from Al 3+ 3 electrons are needed. x moles Al 3+ = (.5 mol e-/4 mol Al) = .11 moles Al 3+ 7. What current is needed to produce 98.5 g of solid gold from Au+ in 2 hours? a. 6.7 A b. 24.1 kA c. 804.0 A d. 7.0 kA I = (Moles of Product)(Moles of electrons)(Faraday)/time I = 0.5 moles * 1 e- *96,485 / (7200 sec) I = 6.7 A 8. The following reaction is at equilibrium. If [Ag + ] = 1x10 -6 M, what is [Cu 2+ ]? Cu 2+ + 2Ag(s) Cu (s) + 2Ag + a. 5x10 -9 M b. 5x10 -15 M c. 2x10 8 M d. 5x10-7 M E o = (0.2) - .8 = -0.6V E = E o - (0.05916/n)*logK 0 = E o - (0.05916/n)*logK -0.6V = (0.05916/n)*log([Ag + ] 2 /[Cu2+]) (1x10 -6 M) 2 /[Cu2+] = 10^(2*-0.6/0.05916) 9. Which of the following is not true about nickel-metal hydride batteries? a.
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e3ps10key - (Hint: ln 2 = 0.6931. ~= 0.7) 1. Balance the...

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