OX # Of N +3 ( ox H. 0)ok. # Of ut 3(-2)=0Ox. # of N= t6Ox. H of O : -2b. P,O.oX. # Of 4(P) + 6(ox.# of 0)= 0OK, of y8) + 6(-2) 20ox. #of P= +3ox # Of O : -2C.Fe,O,OX, # of a(Fe) t 4(ox. #0)=0ox. # of d(Fe)+ 4(-240OX. # of Fe+4Ox # of 0 : -22. Detemine what substance was being oxidized, the substance being reduced and indicatethe oxidation state of all atoms before and after the reaction.2020)2H,0a. CHe +b. Zn2HCI3. Specify which of the following are oxidation-reduction reactions, identify the oxidizing agent,the reducing agent, the substance being oxidized, and the substance being reduced.a. CH, + H,O → CÓ + 3H,b. 2AGNO, + Cu → Cu(NO,), + 2Agc. Zn + 2HCI → ZnCl, + H2d. 2H* + 2Cro, → Cr,07¯ + H,O

Answered: 2. Determine what substance was being…

2. Determine what substance was being oxidized, the substance being reduced and indicate the oxidation state of all atoms before and after the reaction. 2H,O a. CH + 20z) b. Zn 2HCI ZnClaa) +

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

OX # Of N +3 ( ox H. 0)
ok. # Of ut 3(-2)=0
Ox. # of N= t6
Ox. H of O : -2
b. P,O.
oX. # Of 4(P) + 6(ox.# of 0)= 0
OK, of y8) + 6(-2) 20
ox. #of P= +3
ox # Of O : -2
C.
Fe,O,
OX, # of a(Fe) t 4(ox. #0)=0
ox. # of d(Fe)+ 4(-240
OX. # of Fe+4
Ox # of 0 : -2
2. Detemine what substance was being oxidized, the substance being reduced and indicate
the oxidation state of all atoms before and after the reaction.
2020)
2H,0
a. CHe +
b. Zn
2HCI
3. Specify which of the following are oxidation-reduction reactions, identify the oxidizing agent,
the reducing agent, the substance being oxidized, and the substance being reduced.
a. CH, + H,O → CÓ + 3H,
b. 2AGNO, + Cu → Cu(NO,), + 2Ag
c. Zn + 2HCI → ZnCl, + H2
d. 2H* + 2Cro, → Cr,07¯ + H,O

Expert Solution

Step 1

Solving Ques (2)

Oxidation : Increase in +ve oxidation state

Reduction : Decrease in +ve oxidation state

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