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Aluminum forms a tough oxide layer that protects the metal underneath from abrasion and further oxidation. A thin layer of this oxide forms naturally when in contact with air, but with electrochemical anodization one can produce a thicker layer that can significantly extend the lifetime of metal parts. The main form of aluminum oxide is Al2O3, which is a non-conductive network solid. It forms from aluminum via the following series of reactions: Al₂O3 (s) + 6 H+ (aq) + 6 e¯ = 2 Al (s) + 3 H₂0 (1) H₂ (g) = 2 H+ (aq) + 2 e¯ Eº = 0.00 V When exposed to a strong base such as sodium hydroxide, the aluminum oxide can etch away, dissolving: Al₂O3 (s) + 2OH(aq) + 3 H₂O (1) → 2 Al(OH) (aq) You will perform the anodization under acidic conditions, using concentrated sulfuric acid (approximately 18 M) as the electrolyte. Under acidic anodization conditions, aluminum forms first a "barrier layer" of dense oxide coating the electrode, then a thicker and more porous oxide coating (thickness depends on the duration of anodization) [1]. Eº = -1.55 V 3. Reaction studied: Based on the approximate concentration of H* in concentrated sulphuric acid, and assuming 0.212 M of H₂(g) initially present in the acid (based on typical atmospheric compositions), determine the Ecell expected under these conditions. Using the reaction on the top how would u solve question 3. Using the equations ecathode=ecathode- 0.0562(log(k))/ne and the similar equation for eanode the reaction must have H2 gas as a product

Aluminum forms a tough oxide layer that protects the metal underneath from abrasion and further oxidation. A thin layer of this oxide forms naturally when in contact with air, but with electrochemical anodization one can produce a thicker layer that can significantly extend the lifetime of metal parts. The main form of aluminum oxide is Al2O3, which is a non-conductive network solid. It forms from aluminum via the following series of reactions: Al₂O3 (s) + 6 H+ (aq) + 6 e¯ = 2 Al (s) + 3 H₂0 (1) H₂ (g) = 2 H+ (aq) + 2 e¯ Eº = 0.00 V When exposed to a strong base such as sodium hydroxide, the aluminum oxide can etch away, dissolving: Al₂O3 (s) + 2OH(aq) + 3 H₂O (1) → 2 Al(OH) (aq) You will perform the anodization under acidic conditions, using concentrated sulfuric acid (approximately 18 M) as the electrolyte. Under acidic anodization conditions, aluminum forms first a "barrier layer" of dense oxide coating the electrode, then a thicker and more porous oxide coating (thickness depends on the duration of anodization) [1]. Eº = -1.55 V 3. Reaction studied: Based on the approximate concentration of H* in concentrated sulphuric acid, and assuming 0.212 M of H₂(g) initially present in the acid (based on typical atmospheric compositions), determine the Ecell expected under these conditions. Using the reaction on the top how would u solve question 3. Using the equations ecathode=ecathode- 0.0562(log(k))/ne and the similar equation for eanode the reaction must have H2 gas as a product

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Aluminum forms a tough oxide layer that protects the metal underneath from abrasion and further oxidation. A thin layer of this oxide forms naturally when in contact with air, but with electrochemical anodization one can produce a thicker layer that can significantly extend the lifetime of metal parts. The main form of aluminum oxide is Al2O3, which is a non-conductive network solid. It forms from aluminum via the following series of reactions: Al₂O3 (s) + 6 H+ (aq) + 6 e¯ ⇒ 2 Al (s) + 3 H₂0 (1) H₂ (g) 2 H+ (aq) + 2 e¯ Eº = 0.00 V When exposed to a strong base such as sodium hydroxide, the aluminum oxide can etch away, dissolving: Al₂O3 (s) + 2OH(aq) + 3 H₂O (1) → 2 Al(OH) (aq) You will perform the anodization under acidic conditions, using concentrated sulfuric acid (approximately 18 M) as the electrolyte. Under acidic anodization conditions, aluminum forms first a "barrier layer" of dense oxide coating the electrode, then a thicker and more porous oxide coating (thickness depends on the duration of anodization) [1]. Eº = -1.55 V 3. Reaction studied: Based on the approximate concentration of H* in concentrated sulphuric acid, and assuming 0.212 M of H₂(g) initially present in the acid (based on typical atmospheric compositions), determine the Ecell expected under these conditions. Using the reaction on the top how would u solve question 3. Using the equations ecathode=ecathode- 0.0562(log(k))/ne and the similar equation for eanode the reaction must have H2 gas as a product
Transcribed Image Text:Aluminum forms a tough oxide layer that protects the metal underneath from abrasion and further oxidation. A thin layer of this oxide forms naturally when in contact with air, but with electrochemical anodization one can produce a thicker layer that can significantly extend the lifetime of metal parts. The main form of aluminum oxide is Al2O3, which is a non-conductive network solid. It forms from aluminum via the following series of reactions: Al₂O3 (s) + 6 H+ (aq) + 6 e¯ ⇒ 2 Al (s) + 3 H₂0 (1) H₂ (g) 2 H+ (aq) + 2 e¯ Eº = 0.00 V When exposed to a strong base such as sodium hydroxide, the aluminum oxide can etch away, dissolving: Al₂O3 (s) + 2OH(aq) + 3 H₂O (1) → 2 Al(OH) (aq) You will perform the anodization under acidic conditions, using concentrated sulfuric acid (approximately 18 M) as the electrolyte. Under acidic anodization conditions, aluminum forms first a "barrier layer" of dense oxide coating the electrode, then a thicker and more porous oxide coating (thickness depends on the duration of anodization) [1]. Eº = -1.55 V 3. Reaction studied: Based on the approximate concentration of H* in concentrated sulphuric acid, and assuming 0.212 M of H₂(g) initially present in the acid (based on typical atmospheric compositions), determine the Ecell expected under these conditions. Using the reaction on the top how would u solve question 3. Using the equations ecathode=ecathode- 0.0562(log(k))/ne and the similar equation for eanode the reaction must have H2 gas as a product
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