Cr 2 O 7 2 I 2 reducing agents I Cr 3 SO 4 2 2150 reducing agents rxn A B A s B

Cr 2 o 7 2 i 2 reducing agents i cr 3 so 4 2 2150

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Cr 2 O 7 2 - I 2 ; reducing agents: I - Cr 3 + SO 4 2 - 21.50 reducing agents rxn A B A( s ) + B b + ( aq ) B B( s ) + A a + ( aq ) B H 2 B( s ) + 2H + ( aq ) B B b + ( aq ) + H 2 ( g ) C A A( s ) + C c + ( aq ) C( s ) + A a + ( aq ) C A B C( s ) + 2H + ( aq ) B C c + ( aq ) + H 2 ( g ) (Metal C would cause formation of H 2 gas.) 21.51 a) copper b) oxidizing agent: Cu 2 + ; reducing agent: Fe c) yes d) Cu 2 + ( aq ) + Fe( s ) B Cu( s ) + Fe 2 + ( aq ) e) E ° cell = E ° Cu 2 + - E ° Fe 2 + = 0.34V - ( - 0.44V) = 0.78V 21.52 E cell = - 0.0592V log Q n K and G = - n FE cell a) When Q / K < 1, E cell 0 and G < 0. When Q / K = 1, E cell = 0 and G = 0. When Q / K 1, E cell < 0 and G 0. b) Only when Q / K < 1 will the reaction proceed spontaneously and be able to do work.
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21.53 A( s ) + B + ( aq ) B A + ( aq ) + B( s ) a) [A + ] increases and [B + ] decreases. b) E cell decreases. c) E cell = E ° cell - 0.0592V n log [A ] [B ] + + E cell will equal E ° cell when log [A ] [B ] + + = 0, which will occur when [A ] [B ] + + = 1. d) Yes. E cell will be less than E ° cell when Q ( = [A ] [B ] + + for this problem) is greater than 1. 21.54 a) E cell = E ° cell - 2.303 RT n F log Q E cell = 2.303 RT n F log K - 2.303 RT n F log Q E cell = 2.303 RT n F log K Q = - 2.303 RT n F log K Q If Q/K < 1, E cell will decrease with a decrease in cell temperature. If Q / K 1, E cell will increase (become less negative) with a decrease in cell temperature. b) E cell = E ° cell - 2.303 RT n F log [active ion at anode] [active ion at cathode] E cell will decrease as the concentration of an active ion at the anode increases. c) E cell will increase as the concentration of an active ion at the cathode increases. d) E cell will increase as the pressure of a gaseous reactant in the cathode compartment increases. 21.55 In a concentration cell, the redox reaction proceeds in the direction to achieve concentration equality in the half-cells. cathode rxn: M + ( aq ) + e - B M( s ) anode rxn: M( s ) B M + ( aq ) + e - The more concentrated electrolyte is in the cathode compartment. 21.56 a) Ni( s ) + 2Ag + ( aq ) h 2Ag( s ) + Ni 2 + ( aq ) E ° cell = E ° Ag + - E ° Ni 2 + = 0.80V - ( - 0.25V) = 1.05V E ° cell = 0.0592V n log K log K = cell 0.0592V ο n E = 2 x 1.05V 0.0592V = 35.5 K = 3 x 10 35 b) 2Cr 3 + ( aq ) + 3Fe( s ) h 3Fe 2 + ( aq ) + 2Cr( s ) E ° cell = E ° Cr 3 + - E ° Fe 2 + = - 0.74V - ( - 0.44V) = - 0.30V
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log K = cell 0.0592V ο n E = 6 x ( 0.30V) 0.0592V - = - 30.4 K = 4 x 10 - 31 21.57 a) 2Al( s ) + 3Cd 2 + ( aq ) h 2Al 3 + ( aq ) + 3Cd( s ) E ° cell = E ° Cd 2 + - E ° Al 3 + = - 0.40V - ( - 1.66V) = 1.26V log K = cell 0.0592V ο n E = 6 x 1.26V 0.0592V = 127.7 K = 5 x 10 127 b) I 2 ( s ) + 2Br - ( aq ) h Br 2 ( l ) + 2I - ( aq ) E ° cell = E ° I 2 - E ° Br 2 = 0.53V - (1.07V) = - 0.54V log K = cell 0.0592V ο n E = 2 x ( 0.54V) 0.0592V - = - 18.2 K = 6 x 10 - 19 21.58 a) 2 Ag( s ) + Mn 2 + ( aq ) h 2 Ag + ( aq ) + Mn( s ) E ° cell = E ° Mn 2 + - E ° Ag + = - 1.18V - (0.80V) = - 1.98V log K = cell 0.0592V ο n E = 2 x ( 1.98V) 0.0592 - = - 66.9 K = 1 x 10 - 67 b) Cl 2 ( g ) + 2 Br - ( aq ) h Br 2 ( l ) + 2 Cl - ( aq ) E ° cell = E ° Cl 2 - E ° Br 2 = 1.36V - (1.07V) = 0.29V log K = cell 0.0592V ο n E = 2 x (0.29V) 0.0592 = 9.8 K = 6 x 10 9 21.59 a) 2 Cr( s ) + 3 Cu 2 + ( aq ) h 2 Cr 3 + ( aq ) + 3 Cu( s ) E ° cell = E ° Cu 2 + - E ° Cr 3 + = 0.34V - ( - 0.74V) = 1.08V log K = cell 0.0592V ο n E = 6 x (1.08V) 0.0592 = 109.5 K = 3 x 10 109
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b) Sn( s ) + Pb 2 + ( aq ) h Sn 2 + ( aq ) + Pb( s ) E ° = E ° Pb 2 + - E ° Sn 2 + = - 0.13V - ( - 0.14V) = 0.01V log K = cell 0.0592V ο n E = 2 x (0.01V) 0.0592 = 0.3 K = 2 21.60 a) G ° = - nFE ° cell = - (2 mol e - ) x (96.5 kJ/V mol e - ) x (1.05V) = - 203 kJ b) G ° = - nFE ° cell = - (6 mol e - ) x (96.5 kJ/V mol e - ) x ( - 0.30V) = 1.7 x 10 2 kJ 21.61 a) G ° = - nFE ° cell = - (6 mol e - ) x (96.5 kJ/V mol e - ) x (1.26V) = 7.30 x 10 2 kJ b) G ° = - nFE ° cell = - (2 mol e - ) x (96.5 kJ/V mol e - ) x ( - 0.54V) = 1.0 x 10 2 kJ 21.62 a) G ° = - nFE ° cell = - (2 mol e - ) x (96.5 kJ/V mol e - ) x
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