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Unformatted text preview: CHEM 1000 A, V Midterm Test #3, February 3, 2006. Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each) 1. What is the difference between electrowinning and electrorefining? In electrowinning, the metal is simply reduced out of solution. In electrorefining, the metal is first oxidized from the solid anode into solution, then re-reduced at the cathode. 2. Write the shorthand cell notation for the REDOX reaction: 2 Al (s) + 3 H 2 O 2(aq) 2 Al +3 (aq) + 3 O 2(g) + 6 H + (aq) Silly question there is no reduction taking place. 5/5 for everyone! 3. Referring to the diagram of a galvanic cell shown below which has the overall reaction: 5 Cu (s) + 2 MnO 4- (aq) + 16 H + (aq) 5 Cu +2 (aq) + 2 Mn +2 (aq) + 8 H 2 O (l) (i) Which electrode is A (anode or cathode)? Anode (ii) What is electrode A made of? Cu (s) (iii) What is electrode B made of? Pt (s) or other inert metal (iv) Name all species (except spectator ions) in solution C. Cu +2 (aq) (v) Name all species (except spectator ions) in solution D. MnO 4- (aq) , H + (aq) , Mn +2 (aq) 1 D 4. In an electrolytic cell having two Platinum electrodes and Pb(NO 3 ) 2(aq) as an electrolyte, what is produced at the anode and what is produced at the cathode? (Be careful there are three reducible species in this solution!) At the anode: Pb +2 cannot be further oxidized. NO 3- (aq) cannot be further oxidized (note that the N atom has an oxidation number of +5, which is as high as it can be) Thus, only the water can be oxidized: 2 H 2 O (l) O 2(g) + 4 H + (aq) + 4 e, and so O 2(g) and H + (aq) are produced at the anode. At the cathode: Pb +2 + 2 e Pb (s) E = -0.13 V 2 H 2 O + 2 e H 2(g) + 2 OH (aq) E = -0.83 V NO 3- (aq) + 4 H + (aq) + 3 e- NO (g) + 2 H 2 O (l) E = +0.96 V Reduction of the NO 3- (aq) has the most positive potential, so NO (g) is produced at the cathode....
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