(a)
Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.
Zn(s)+ 2 HNO3(aq)→ Zn(NO3)2(aq)+ H2(g)
Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.
Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.
(a)
Answer to Problem 18A
- H= Reduce.
- Zn = Oxidize.
Explanation of Solution
The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.
Zn(s)+ 2 HNO3(aq)→ Zn(NO3)2(aq)+ H2(g)
The oxidation number of elements are:
Zn(s)+ 2 HNO3(aq)→ Zn(NO3)2(aq)+ H2(g) 0 +1+5-2 +2 +5 -2 0
In the given reaction, the oxidation number of hydrogen in HNO3 changes from +1 to 0 so it is a reduction process. Hence hydrogen in HNO3 reduces during reaction.
Zinc in Zn oxidizes from 0 to +2, thus, Zn oxidized during the given reaction.
(b)
Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.
H2(g)+ CuSO4(aq)→Cu(s)+ H2SO4(aq)
Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.
Oxidation and reduction reactions are the processes involved in loss or gain of electron.
Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.
(b)
Answer to Problem 18A
- Cu = Reduce.
- H = Oxidize.
Explanation of Solution
The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.
H2(g)+ CuSO4(aq)→Cu(s)+ H2SO4(aq)
The oxidation number of elements are:
H2(g)+ CuSO4(aq)→ Cu(s)+ H2SO4(aq) 0 +2+6-2 0 +1+6-2
In the given reaction, the oxidation number ofCu in CuSO4(aq) changes from +2 to 0 so it is a reduction process. Hence Cuin CuSO4(aq) reduces during reaction.
Hydrogen in H2 oxidizes from 0 to +1, thus, H2 oxidized during the given reaction.
(c)
Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.
N2(g)+ 3 Br2(l)→ 2 NBr3(g)
Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.
Oxidation and reduction reactions are the processes involved in loss or gain of electron.
Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.
(c)
Answer to Problem 18A
- Br= Reduce.
- N = Oxidize.
Explanation of Solution
The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.
N2(g)+ 3 Br2(l)→ 2 NBr3(g)
The oxidation number of elements are:
N2(g)+ 3 Br2(l)→ 2 NBr3(g) 0 0 +3-1
In the given reaction, the oxidation number ofBr in Br2(l) changes from 0 to -1 so it is a reduction process. Hence Brin Br2(l) reduces during reaction.
Nitrogen in N2 oxidizes from 0 to +3, thus, N2 oxidized during the given reaction.
(d)
Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.
2 KBr(aq)+ Cl2(g)→ 2 KCl(aq) + Br2(l)
Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.
Oxidation and reduction reactions are the processes involved in loss or gain of electron.
Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.
(d)
Answer to Problem 18A
- Br = oxidize.
- Cl = Reduce.
Explanation of Solution
The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.
2 KBr(aq)+ Cl2(g)→ 2 KCl(aq) + Br2(l)
The oxidation number of elements are:
2 KBr(aq)+ Cl2(g)→ 2 KCl(aq) + Br2(l) +1 -1 0 +1-1 0
In the given reaction, the oxidation number ofBr in KBr changes from -1 to 0 so it is a oxidation process. Hence Br in KBr oxidizes during reaction.
Chlorine in Cl2 oxidizes from 0 to -1, thus, Cl2 reduces during the given reaction.
Chapter 18 Solutions
World of Chemistry, 3rd edition
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