Chapter-11-HW

Chapter-11-HW - Homework #3 Chapter 11 Electrochemistry...

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1 Homework #3 Chapter 11 Electrochemistry Chapter 4 63. a) Oxidation ½ Reaction Fe + HCl HFeCl 4 Fe + 4HCl HFeCl 4 Fe + 4HCl HFeCl 4 + 3H + Fe + 4HCl HFeCl 4 + 3H + + 3e - Reduction ½ Reaction H 2 2H + H 2 2H + + 2e- H 2 Balanced Reaction 2(Fe + 4HCl HFeCl 4 + 3H + + 3e - ) 3(2H + + 2e- H 2 ) 2Fe(s) + 8HCl(aq) 2HFeCl 4 (aq) + 3H 2 (g) b) Oxidization ½ Reaction I - I 3 - 3I - I 3 - 3I - I 3 - + 2e - Reduction ½ Reaction IO 3 - I 3 - 3IO 3 - I 3 - 3IO 3 - I 3 - + 9H 2 O 3IO 3 - + 18H + I 3 - + 9H 2 O 3IO 3 - + 18H + + 16e - I 3 - + 9H 2 O Balanced Reaction 3IO 3 - + 18H + + 16e - I 3 - + 9H 2 O 8(3I - I 3 - + 2e - ) 3IO 3 - (aq) + 18H + (aq) + 24I - (aq) 9I 3 - (aq) + 9H 2 O(l) Divide trough by 3 IO 3 - (aq) + 6H + (aq) + 8I - (aq) 3I 3 - (aq) + 3H 2 O(l) c) Oxidation ½ Reaction Cr(NCS) 6 4- Cr 3+ + NO 3 - + CO 2 + SO 4 2- Cr(NCS) 6 4- Cr 3+ + 6NO 3 - + 6CO 2 + 6SO 4 2- Cr(NCS) 6 4- + 54H 2 O Cr 3+ + 6NO 3 - + 6CO 2 + 6SO 4 2- Cr(NCS) 6 4- + 54H 2 O Cr 3+ + 6NO 3 - + 6CO 2 + 6SO 4 2- + 108H + Cr(NCS) 6 4- + 54H 2 O Cr 3+ + 6NO 3 - + 6CO 2 + 6SO 4 2- + 108H + + 97e -
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2 Reduction ½ Reaction Ce 4+ Ce 3+ Ce 4+ + e - Ce 3+ Balanced Reaction Cr(NCS) 6 4- + 54H 2 O Cr 3+ + 6NO 3 - + 6CO 2 + 6SO 4 2- + 108H + + 97e - 97(Ce 4+ + e - Ce 3+ ) Cr(NCS) 6 4- (aq)+ 54H 2 O(l) + 97Ce 4+ (aq) Cr 3+ (aq) + 6NO 3 - (aq) + 6CO 2 (g) + 6SO 4 2 (aq) + 108H + (aq) + 97Ce 3+ (aq) d) Oxidation ½ Reaction CrI 3 CrO 4 2- + IO 4 - CrI 3 CrO 4 2- + 3IO 4 - CrI 3 + 16H 2 O CrO 4 2- + 3IO 4 - CrI 3 + 16H 2 O CrO 4 2- + 3IO 4 - + 32H + CrI 3 + 16H 2 O CrO 4 2- + 3IO 4 - + 32H + + 27e - Reduction ½ Reaction Cl 2 Cl - Cl 2 2Cl - Cl 2 + 2e - 2Cl - Balanced Equation 2(CrI 3 + 16H 2 O CrO 4 2- + 3IO 4 - + 32H + + 27e - ) 27(Cl 2 + 2e - 2Cl - ) 2CrI 3 (s) + 32H 2 O(l) + 27Cl 2 (g) 2CrO 4 2- (aq) + 6IO 4 - (aq) + 64H + (aq) + 54Cl - (aq) The solution is basic not acidic, add 64 OH - to both sides 2CrI 3 (s) + 27Cl 2 (g) + 64OH - (aq) 2CrO 4 2- (aq) + 6IO 4 - (aq) + 32H 2 O(l) + 54Cl - (aq) e) Oxidation ½ Reaction Fe(CN) 6 4- Fe(OH) 3 + CO 3 2- + NO 3 - Fe(CN) 6 4- Fe(OH) 3 + 6CO 3 2- + 6NO 3 - Fe(CN) 6 4- + 39H 2 O Fe(OH) 3 + 6CO 3 2- + 6NO 3 - Fe(CN) 6 4- + 39H 2 O Fe(OH) 3 + 6CO 3 2- + 6NO 3 - + 75H + Fe(CN) 6 4- + 39H 2 O Fe(OH) 3 + 6CO 3 2- + 6NO 3 - + 75H + + 61e - Reduction ½ Reaction Ce 4+ Ce(OH) 3 Ce 4+ + 3H 2 O Ce(OH) 3 Ce 4+ + 3H 2 O Ce(OH) 3 + 3H + Ce 4+ + 3H 2 O + e - Ce(OH) 3 + 3H + Balance Reaction Fe(CN) 6 4- + 39H 2 O Fe(OH) 3 + 6CO 3 2- + 6NO 3 - + 75H + + 61e - 61(Ce 4+ + 3H 2 O + e - Ce(OH) 3 + 3H + ) Fe(CN) 6 4- (aq) + 222H 2 O(l) + 61Ce 4+ (aq) Fe(OH) 3 (s) + 6CO 3 2- (aq) + 6NO 3 - (aq)+ 258H + (aq) + 61Ce(OH) 3 (s)
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3 The solution is basic not acidic, add 258OH - to both sides Fe(CN) 6 4- (aq) + 258OH - (aq) + 61Ce 4+ (aq) Fe(OH) 3 (s) + 6CO 3 2- (aq) + 6NO 3 - (aq)+ 61Ce(OH) 3 (s) + 36H 2 O(l) Chapter 11 2. Reactions of Interest Ni 2+ + 2e - Ni E ° =-0.23 V Cu + + e - Cu E ° =0.52 V Cu 2+ + 2e - Cu E ° =0.34 V Zn 2+ + 2e - Zn E ° =-0.76 V In order to plate out Ni you need the Ni reaction to be the reduction ½ reaction (cathode). In addition, you also need the cell to be galvanic (E ° >0). The E ° cell of the nickel/copper cell is -0.75 V or -0.57 V depending on the ion of copper that is used. Therefore, neither of these cells would be a galvanic cell, resulting in copper not being an appropriate material. The E ° cell of the nickel/zinc cell is 0.53 V. Therefore, the nickel/zink cell would plate out Ni.
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Chapter-11-HW - Homework #3 Chapter 11 Electrochemistry...

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