In the oxidation of aluminium, the electrode reactions are as follows: C a t h o d e ( r e d u c t i o n ) : 6 H + ( a q ) + 6 e − → 3 H 2 ( g ) A n o d e ( O x i d a t i o n ) : 1 A l ( s ) + 3 H 2 0 ( l ) → A l 2 0 3 ( s ) + 6 H + ( a q ) + 6 e − _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ O v e r a l l r e a c t i o n : 2 A l ( s ) + 2 H 2 0 ( l ) → A l 2 0 3 ( s ) + 3 H 2 ( g ) The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis . Find the minutes required to produce a 0.0100 m m thick coating of A l 2 O 3 , having density as 3.97 g / c m 3 on a square piece of aluminium metal 10.0 c m on an edge if the current passed through the piece is 0.600 A Concept Introduction: Calculate the mass of A l 2 O 3 from its density and volume of the coating first. Then find the moles of A l 2 O 3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of A l 2 O 3 and mole ratio of A l 2 O 3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of A l 2 O 3 . Given: The electrode reactions in the oxidation of aluminium, C a t h o d e ( r e d u c t i o n ) : 6 H + ( a q ) + 6 e − → 3 H 2 ( g ) A n o d e ( O x i d a t i o n ) : 1 A l ( s ) + 3 H 2 0 ( l ) → A l 2 0 3 ( s ) + 6 H + ( a q ) + 6 e − _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ O v e r a l l r e a c t i o n : 2 A l ( s ) + 2 H 2 0 ( l ) → A l 2 0 3 ( s ) + 3 H 2 ( g ) A l 2 O 3 has density of 3.97 g / c m 3 and current passed through the aluminium piece is 0.600 A

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Answer 1

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Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

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Answer 2

Question

Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

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Answer 3

Question

Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

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Answer 4

Question

Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

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Answer 5

Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

Answer 6

Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

Answer 7

Chapter 18, Problem 18.134CP

Interpretation Introduction

Interpretation:

In the oxidation of aluminium, the electrode reactions are as follows:

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

The thickness of the aluminium oxide coating that forms an anode can be conrolled by varying the current flow during the electrolysis. Find the minutes required to produce a 0.0100 mm thick coating of Al2O3 , having density as 3.97 g/cm3 on a square piece of aluminium metal 10.0 cm on an edge if the current passed through the piece is 0.600 A

Concept Introduction: Calculate the mass of Al2O3 from its density and volume of the coating first. Then find the moles of Al2O3 from its calculated mass and molar mass. Then find the number of moles of electrons from the moles of Al2O3 and mole ratio of Al2O3 to electrons, followed by calculating the charge from the moles of electron. At the end, find out the time required for coating of Al2O3.

Given: The electrode reactions in the oxidation of aluminium,

Cathode ( reduction): 6H+( aq) + 6 e−→3H2(g)Anode ( Oxidation): 1Al(s)+3H20(l)→ Al203(s)+6H+( aq)+6 e−Overall reaction: 2Al(s)+2H20(l) →Al203(s) + 3H2(g)

Al2O3 has density of 3.97 g/cm3 and current passed through the aluminium piece is 0.600 A

Answer 8

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Answer 11

Ch. 18 - Prob. 18.1P Ch. 18 - Prob. 18.2A Ch. 18 - Prob. 18.3P Ch. 18 - APPLY 18.4 Balance the following net ionic...Ch. 18 - Prob. 18.5P Ch. 18 - Prob. 18.6A Ch. 18 - PRACTICE 18.7 Write a balanced equation for the...Ch. 18 - Prob. 18.8A Ch. 18 - Prob. 18.9P Ch. 18 - Prob. 18.10A

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