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- The following graph represents the Relation between time and mass of Precipitate copper when passing Electricity in Cuso4 solution using inert Electrodes. Mass of copper (8) 0.50 0.40 0.30 0.20 0.10 T(sec) 10 20 30 40 Study the graph well then answer: - The anode half reaction is ....... (а) 2H* + 2e → H2 (g) (aq) (b) 4H* (aq) + 4e → H2 (g) (C) 2Cu++ (aq) + 4e → 2Cu (s) (d) Cu*+ (ag) + 2e → Cu (s) a b

- The following graph represents the Relation between time and mass of Precipitate copper when passing Electricity in Cuso4 solution using inert Electrodes. Mass of copper (8) 0.50 0.40 0.30 0.20 0.10 T(sec) 10 20 30 40 Study the graph well then answer: - The anode half reaction is ....... (а) 2H* + 2e → H2 (g) (aq) (b) 4H* (aq) + 4e → H2 (g) (C) 2Cu++ (aq) + 4e → 2Cu (s) (d) Cu*+ (ag) + 2e → Cu (s) a b

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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- The following graph represents the Relation between time and mass of Precipitate copper when passing Electricity in CuSO4 solution using inert Electrodes. Mass of copper (g) 0.50 0.40 0.30 0.20 0.10 T(sec) 10 20 30 40 Study the graph well then answer: - The anode half reaction is (a) 2H* (aq) + 2e → H2 (g) (b) 4H* (aq) + 4e → H2 (g) (C) 2Cu** (aq) + 4e → 2Cu (s) (d) Cu+ (aq) + 2e → Cu (s) a b C II
Transcribed Image Text:- The following graph represents the Relation between time and mass of Precipitate copper when passing Electricity in CuSO4 solution using inert Electrodes. Mass of copper (g) 0.50 0.40 0.30 0.20 0.10 T(sec) 10 20 30 40 Study the graph well then answer: - The anode half reaction is (a) 2H* (aq) + 2e → H2 (g) (b) 4H* (aq) + 4e → H2 (g) (C) 2Cu** (aq) + 4e → 2Cu (s) (d) Cu+ (aq) + 2e → Cu (s) a b C II
- The anode half reaction is (а) 20H: → 2H+ (aq) + 02 (g) + 4e (аq) (b) 40Н (ag) → 4H* (aq) + 20 2 (g) + 8e (C) so- 4(aq) → SO2 (ag) + O2 (g) + 2e (d) 2S0 4(aq) → 2S02 (ag) + 02 (g) + 4e a b d - The electric current intensity passed in the electrolyte equal. (а) 20A (b) 40A (C) 50A (d) 60A II
Transcribed Image Text:- The anode half reaction is (а) 20H: → 2H+ (aq) + 02 (g) + 4e (аq) (b) 40Н (ag) → 4H* (aq) + 20 2 (g) + 8e (C) so- 4(aq) → SO2 (ag) + O2 (g) + 2e (d) 2S0 4(aq) → 2S02 (ag) + 02 (g) + 4e a b d - The electric current intensity passed in the electrolyte equal. (а) 20A (b) 40A (C) 50A (d) 60A II
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