- 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
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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Question
Answer my question please
![- 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a921efd-504e-4713-80c4-084ce0999f2c%2F6caff0f7-1288-480c-86e7-8b9907f32b6f%2Fugmq49g_processed.jpeg&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a921efd-504e-4713-80c4-084ce0999f2c%2F6caff0f7-1288-480c-86e7-8b9907f32b6f%2Fhhvgb4_processed.jpeg&w=3840&q=75)
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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