P4(s) + OH(aq) - PH3 (g) + HPO(aq)
Answers
Explanation:
प्रकृति में ऊर्जा कई अलग अलग रूपों में मौजूद है। इन के उदाहरण हैं: प्रकाश ऊर्जा, यांत्रिक ऊर्जा, गुरुत्वाकर्षण ऊर्जा, विद्युत ऊर्जा, ध्वनि ऊर्जा, रसायनिक ऊर्जा और परमाणु ऊर्जा। प्रत्येक ऊर्जा को एक अन्य रूप में परिवर्तित या बदला जा सकता है। ऊर्जा के कई विशिष्ट प्रकारों में प्रमुख रूप गतिज ऊर्जा और स्थितिज ऊर्जा है।
1 A
1 B
1 C
1 D
1 E
1 F
1 G
1 H
2 A
2 B
2 C
2 D
2 E
3 A
3 B
3 C
3 D
3 E
4
5
6 A
6 B
6 C
6 D
6 E
6 F
7
8
9
10
11
12 A
12 B
13 A
13 B
13 C
13 D
13 E
14
15
16
17
18 A
18 B
18 C
18 D
19 A
19 B
19 C
20
21
22
23
24
25
26 A
26 B
26 C
26 D
26 E
27
28
29
30
Q. 19 A
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
P4(s) + OH–(aq) → PH3(g) + HPO–2(aq)
Class 11th
NCERT - Chemistry Part-II
8. Redox reactions
Answer :
The O.N. (oxidation number) of P decreases from 0 in P4 to – 3 in PH3 and increases from 0 in P4 to + 2 in HPO2-.
Hence, P4 clearly acts both as an oxidizing agent and a reducing agent in this reaction.
Now, balancing the equation in basic medium by ion-electron method for the reduction half reaction: -
0P4 (s) → -3PH3(g)
Now, balancing P atoms :-
P4 (s) → 4PH3(g)
Balancing oxidation number by adding electrons: -
P4 (s) +12 e-→ 4PH3(g)
Balancing charge by adding OH- ions : -
P4 (s) +12 e-→ 4PH3(g) +12OH-1(aq)
Balancing 'O' atoms by adding H2O : -
Balancing the equation in basic medium by ion-electron method for the oxidation half reaction: -
Now, balancing P atoms : -
Balance oxidation number by adding electrons : -
Balance charge by adding OH- ions we get :-
Oxygen and hydrogen are balanced automatically.
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction equation as:-
Oxidation number method