Answers_to_assn_3 - CHEM 123 Assingment#3 Answers page 1 of...

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CHEM 123: Assingment #3 Answers page 1 of 2 1. Given that E º = +0.04 V for the reaction: Cd 2 + + Fe(s) Fe 2 + + Cd(s) what is the standard reduction potential for Cd 2 + + 2e - Cd(s)? (Look up E º for the Fe 2 + /Fe half reaction.) Oxidation: Fe(s) Fe 2+ (aq) + 2e - o oxid E = +0.41 V (from text) Reduction: Cd 2+ (aq) + 2e - Cd(s) o red E = ? Overall: Cd 2 + + Fe(s) Fe 2 + + Cd(s) o o o tot oxid red 0.04 V E E E = + = o red E = 0.04 V - 0.41 V = - 0.37 V (Note: This value does not agree exactly with the E o given in the text.) 2. (a) What is E o for each of these reactions? (b) In which direction does each reaction proceed spontaneously at standard state conditions? (c) What is the equilibrium constant in each case? (1) Pb(s) + Co 2 + Pb 2 + + Co(s) (2) PbSO 4 (s) + Ni(s) Ni 2 + + Pb(s) + SO 4 2 - (3) IO 3 - + 6H + + 5Ag(s) ½ I 2 (s) + 3H 2 O + 5Ag + Values form Text book 2 3 IO - (aq) + 12H + (aq) + 10e - I 2 (s) + 6H 2 O(l) 1.209 V 1.20 Ag + (aq) + e - Ag(s) 0.799 V 0.800 Pb 2+ (aq) + 2e - Pb(s) - 0.127 V -0.125 Ni 2+ (aq) + 2e - Ni(s) - 0.236 V -0.257 Co 2+ (aq) + 2e - Co(s) - 0.282 V -0.277 PbSO 4 (s) + 2e - Pb(s) + 2 4 SO - - 0.356 V -0.403 not in book For reaction (1): Oxidation: Pb(s) Pb 2+ (aq) + 2e - o
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