1. Assign oxidation no. to the elements underlined:
(i) NaH2P––O4
(ii) NaHS––O4
(iii) H4P––2O7
(iv) K2Mn––––O4
(v) CaO––2
(vi) NaB––H4
(vii) H2S––2O7
(viii) KAl(S––O4)2.12H2O
Answer:
(i) NaH2P––O4
Let x be the oxidation no. of P.
Oxidation no. of Na = +1
Oxidation no. of H = +1
Oxidation no. of O = -2
1.1a
Then,
1(+1) + 2(+1) + 1(x) + 4(-2) = 0
1 + 2 + x -8 = 0
x = +5
Therefore, oxidation no. of P is +5.
(ii) NaHS––O4
1.2a
Let x be the oxidation no. of S.
Oxidation no. of Na = +1
Oxidation no. of H = +1
Oxidation no. of O = -2
Then,
1(+1) + 1(+1) + 1(x) + 4(-2) = 0
1 + 1 + x -8 = 0
x = +6
Therefore, oxidation no. of S is +6.
(iii) H4P––2O7
1.3a
Let x be the oxidation no. of P.
Oxidation no. of H = +1
Oxidation no. of O = -2
Then,
4(+1) + 2(x) + 7(-2) = 0
4 + 2x – 14 = 0
2x = +10
x = +5
Therefore, oxidation no. of P is +5.
(iv) K2Mn––––O4
1.4a
Let x be the oxidation no. of Mn.
Oxidation no. of K = +1
Oxidation no. of O = -2
Then,
2(+1) + x + 4(-2) = 0
2 + x – 8 = 0
x = +6
Therefore, oxidation no. of Mn is +6.
(v) CaO––2
1.5a
Let x be the oxidation no. of O.
Oxidation no. of Ca = +2
Then,
(+2) + 2(x) = 0
2 + 2x = 0
2x = -2
x = -1
Therefore, oxidation no. of O is -1.
(vi) NaB––H4
1.6a
Let x be the oxidation no. of B.
Oxidation no. of Na = +1
Oxidation no. of H = -1
Then,
1(+1) + 1(x) + 4(-1) = 0
1 + x -4 = 0
x = +3
Therefore, oxidation no. of B is +3.
(vii) H2S––2O7
1.7a
Let x be the oxidation no. of S.
Oxidation no. of H = +1
Oxidation no. of O = -2
Then,
2(+1) + 2(x) + 7(-2) = 0
2 + 2x – 14 = 0
2x = +12
x = +6
Therefore, oxidation no. of S is +6.
(viii) KAl(S––O4)2.12H2O
1.8a
Let x be the oxidation no. of S.
Oxidation no. of K = +1
Oxidation no. of Al = +3
Oxidation no. of O= -2
Oxidation no. of H = +1
Then,
1(+1) + 1(+3) + 2(x) + 8(-2) + 24(+1) + 12(-2) = 0
1 + 3 + 2x – 16 + 24 – 24 = 0
2x = +12
x = +6
Therefore, oxidation no. of S is +6.
OR
Ignore the water molecules because it is neutral. Then, the summation of the oxidation no. of all atoms of water molecules can be taken as 0. Hence, ignore the water molecule.
1(+1) + 1(+3) + 2(x) + 8(-2) = 0
1 + 3 + 2x -16 = 0
2x = 12
x = +6
Therefore, oxidation no. of S is +6.
2.Assign oxidation no. to the elements underlined:
(i) KI–3
(ii) H2S––4O6
(iii) Fe–––3O4
(iv) C––H3C––H2OH
(v) C––H3C––OOH
Answer:
(i) KI–3
Let x be the oxidation no. of I.
Oxidation no. of K = +1
Then,
1(+1) + 3(x) = 0
1 + 3x = 0
x = −13
Oxidation no. cannot be fractional. Hence, consider the structure of KI3.
In KI3 molecule, an iodine atom forms coordinate covalent bond with an iodine molecule.
2.1a
Therefore, in KI3 molecule, the oxidation no. of I atoms forming the molecule I2 is 0, while the oxidation no. of I atom which is forming coordinate bond is -1.
(ii) H2S––4O6
2.2a
Let x be the oxidation no. of S.
Oxidation no. of H = +1
Oxidation no. of O = -2
Then,
2(+1) + 4(x) + 6(-2) = 0
2 + 4x -12 = 0
4x = 10
x = +212
Oxidation no. cannot be fractional. Therefore, S would be present with different oxidation state in molecule.
2.21a
The oxidation no. of two out of the four S atoms is +5 while that of other two atoms is 0.
(iii) Fe–––3O4
Let x be the oxidation no. of Fe.
Oxidation no. of O = -2
Then,
3(x) + 4(-2) = 0
3x -8 = 0
x = 83
Oxidation no. cannot be fractional.
One of the three atoms of Fe has oxidation no. +2 and other two atoms of Fe has oxidation no. +3.
2.3a
(iv) C––H3C––H2OH
2.4a
Let x be the oxidation no. of C.
Oxidation no. of O= -2
Oxidation no. of H = +1
Then,
2(x) + 4(+1) + 1(-2) = 0
2x + 4 -2 = 0
x = -2
Therefore, oxidation no. of C is -2.
(v) C––H3C––OOH
2.5a
Let x be the oxidation no. of C.
Oxidation no. of O= -2
Oxidation no. of H = +1
Then,
2(x) + 4(+1) + 2(-2) = 0
2x + 4 -4 = 0
x = 0
Therefore, average oxidation no. of C is 0. Both the carbon atoms are present in different environments so they cannot have same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in CH3COOH.
2.51a
3. The reactions given below are redox reactions. Justify the reactions.
(i) CuO(s)+H2(g)→Cu(s)+H2O(g)
(ii) Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
(iii)4BCl3(g)+3LiAlH4(s)→2B2H6(g)+3LiCl(s)+3AlCl3(s)
(iv) 2K(s)+F2(g)→2K+F(s)
(v) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
Answer:
(i) CuO(s)+H2(g)→Cu(s)+H2O(g)
Oxidation no. of Cu and O in CuO is +2 and -2 respectively.
Oxidation no. of H2 is 0.
Oxidation no. of Cu is 0.
Oxidation no. of H and O in H2O is +1 and -2 respectively.
The oxidation no. of Cu decreased from +2 in CuO to 0 in Cu. That is CuO is reduced to Cu.
The oxidation no. of H increased from 0 to +1 in H2. That is H2 is oxidized to H2O.
Therefore, the reaction is redox reaction.
(ii) Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
In the above reaction,
Oxidation no. of Fe and O in Fe2O3 is +3 and -2 respectively.
Oxidation no. of C and O in CO is +2 and -2 respectively.
Oxidation no. of Fe is 0.
Oxidation no. of C and O in CO2 is +4 and -2 respectively.
The oxidation no. of Fe decreased from +3 in Fe2O3 to 0 in Fe. That is Fe2O3 is reduced to Fe.
The oxidation no. of C increased from 0 to +2 in CO to +4 in CO2. That is CO is oxidized to CO2.
Therefore, the reaction is redox reaction.
(iii)4BCl3(g)+3LiAlH4(s)→2B2H6(g)+3LiCl(s)+3AlCl3(s)
the above reaction,
Oxidation no. of B and Cl in BCl3 is +3 and -1 respectively.
Oxidation no. of Li, Al and H in LiAlH4 is +1, +3 and -1 respectively.
Oxidation no. of B and H in B2H6 is -3 and +1 respectively.
Oxidation no. of Li and Cl in LiCl is +1 and -1 respectively.
Oxidation no. of Al and Cl in AlCl3 is +3 and -1 respectively.
The oxidation no. of B decreased from +3 in BCl3 to -3 in B2H6. That is BCl3is reduced to B2H6.
The oxidation no. of H increased from -1 in LiAlH4 to +1 in B2H6. That is LiAlH4 is oxidized to B2H6.
Therefore, the reaction is redox reaction.
(iv) 2K(s)+F2(g)→2K+F(s)
In the above reaction,
Oxidation no. of K is 0.
Oxidation no. of F is 0.
Oxidation no. of K and F in KF is +1 and -1 respectively.
The oxidation no. of K increased from 0 in K to +1 in KF. That is K is oxidized to KF.
The oxidation no. of F decreased from 0 in F2 to -1 in KF. That is F2 is reduced to KF.
Therefore, the reaction is a redox reaction.
(v) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
In the above reaction,
Oxidation no. of N and H in NH3 is -3 and +1 respectively.
Oxidation no. of O2 is 0.
Oxidation no. of N and O in NO is +2 and -2 respectively.
Oxidation no. of H and O in H2O is +1 and -2 respectively.
The oxidation no. of N increased from -3 in NH3 to +2 in NO.
The oxidation no. of O2 decreased from 0 in O2 to -2 in NO and H2O. That is O2 is reduced.
Therefore, the reaction is a redox reaction.
4. Give the reaction of fluorine when reacts with ice:
H2O(s)+F2(g)→HF(g)+HOF(g)
Give reason that the above reaction is redox reaction.
Answer:
H2O(s)+F2(g)→HF(g)+HOF(g)
In the above reaction,
Oxidation no. of H and O in H2O is +1 and -2 respectively.
Oxidation no. of F2 is 0.
Oxidation no. of H and F in HF is +1 and -1 respectively.
Oxidation no. of H, O and F in HOF is +1, -2 and +1 respectively.
The oxidation no. of F increased from 0 in F2 to +1 in HOF.
The oxidation no. of F decreased from 0 in O2 to -1 in HF.
Therefore, F is both reduced as well as oxidized. So, it is redox reaction.
5. Calculate the oxidation no. of chromium, sulphur and nitrogen in H2SO5, Cr2O2−7 and NO–3. Give the structure for the compounds. Count for the fallacy.
Answer:
(i) H2SO5
Let x be the oxidation no. of S.
Oxidation no. of O= -2
Oxidation no. of H = +1
5.1a
Then,
2(+1) + 1(x) + 5(-2) = 0
2 + x – 10 = 0
x = +8
But the oxidation no. of S cannot be +8 as S has 6 valence electrons. Therefore, the oxidation no. of S cannot be more than +6.
The structure of H2SO5 is as given below:
Now,
2(+1) + 1(x) + 3(-2) + 2(-1) = 0
2 + x – 6 – 2 = 0
x = +6
Therefore, the oxidation no. of S is +6.
(ii) Cr2O2−7
Let x be the oxidation no. of Cr.
Oxidation no. of O= -2
Then,
2(x) + 7(-2) = -2
2x -14 = -2
x = +6
There is no fallacy about the oxidation no. of Cr in Cr2O2−7.
The structure of Cr2O2−7 is as given below.
Each of the two Cr atoms has the oxidation no. of +6.
5.2a
(iii) NO–3
Let x be the oxidation no. of N.
Oxidation no. of O= -2
Then,
1(x) + 3(-2) = -1
x – 6 = -1
x = +5
There is no fallacy about the oxidation no. of N in NO–3.
The structure of NO–3 is as given below.
5.3a
Nitrogen atom has the oxidation no. of +5.
6.Give the formula for the given compounds:
(i) Mercury
(II) chloride
(ii) Nickel (II) sulphate
(iii) Tin (IV) oxide
(iv) Thallium (I) sulphate
(v) Iron (III) sulphate
(vi) Chromium (III) oxide
Answer:
(i) Mercury (II) chloride
HgCl2
(ii) Nickel (II) sulphate
NiSO4
(iii) Tin (IV) oxide
SnO2
(iv) Thallium (I) sulphate
Tl2SO4
(v) Iron (III) sulphate
Fe2(SO4)3
(vi) Chromium (III) oxide
Cr2O3
7.Give a list of the compounds where carbon has oxidation no. from -4 to +4 and nitrogen from -3 to +5.
Answer:
The compound where carbon has oxidation no. from -4 to +4 is as given below in the table:
Compounds Oxidation no. of carbon
CH2Cl2 0
HC≡CH -1
ClC≡CCl +1
CH3Cl -2
CHCl3, CO +2
H3C−CH3 -3
Cl3C−CCl3 +3
CH4 -4
CCl4, CO2 +4
Compounds Oxidation no. of nitrogen
N2 0
N2H2 -1
N2O +1
N2H4 -2
NO +2
NH3 -3
N2O3 +3
NO2 +4
N2O5 +5
Hydrogen peroxide and sulphur dioxide can act as oxidizing and reducing agents in the reactions, nitric acid and ozone act only as oxidants. Why?
Answer:
In sulphur dioxide (SO2) the oxidation no. of S is +4 and the range of oxidation no. of sulphur is from +6 to -2.
Hence, SO2 can act as reducing and oxidising agent.
In hydrogen peroxide (H2O2) the oxidation no. of O is -1 and the range of the oxidation no. of oxygen is from 0 to -2. Oxygen can sometimes attain the oxidation no. +1 and +2.
Therefore, H2O2 can act as reducing and oxidising agent.
In ozone (O3) the oxidation no. of O is 0 and the range of the oxidation no. of oxygen is from 0 to –2. Hence, the oxidation no. of oxygen only decreases in this case.
Therefore, O3 acts only as an oxidant.
In nitric acid (HNO3) the oxidation no. of nitrogen is +5 and the range of the oxidation no. that nitrogen can have is from +5 to -3. Hence, the oxidation no. of nitrogen can only decrease in this case.
Therefore, HNO3 acts only as an oxidant.
9.Consider the following reactions:
(i) 6CO2(g)+6H2O(l)→C6H12O6(aq)+6O2(g)
(ii) O3(g)+H2O2(l)→H2O(l)+2O2(g)
It is suitable to write the above equations as given below. Why?
(i) 6CO2(g)+12H2O(l)→C6H12O6(aq)+6H2O(l)+6O2(g)
(ii) O3(g)+H2O2(l)→H2O(l)+O2(g)+O2(g)
Give a technique to find the path of above (i) and (ii) redox reactions.
Answer:
(i)
Step 1 :
H2O breaks to give H2 and O2.
2H2O(l)→2H2(g)+O2(g)
Step 2 :
The H2 produced in earlier step reduces CO2, thus produce glucose and water.
6CO2(g)+12H2(g)→C6H12O6(s)+6H2O(l)
The net reaction is as given below:
[2H2O(l)→2H2(g)+O2(g)] × 6
6CO2(g)+12H2(g)→C6H12O6(s)+6H2O(l)
————————————————————————————————————————–
6CO2(g)+12H2O(l)→C6H12O6(g)+6H2O(l)+6O2(g)
This is the suitable way to write the reaction as the reaction also produce water molecules in the photosynthesis process.
The path can be found with the help of radioactive H2O18 instead of H2O.
(ii)
Step 1 :
O2 is produced from each of the reactants O3 and H2O2. That is the reason O2 is written two times.
O3 breaks to form O2 and O.
Step 2 :
H2O2 reacts with O produced in the earlier step, thus produce H2O and O2.
O3(g)→O2(g)+O(g)
H2O2(l)+O(g)→H2O(l)+O2(g)
———————————————————————————————-
H2O2(l)+O3(g)→H2O(l)+O2(g)+O2(g)
The path can be found with the help of H2O182 or O183.
10. AgF2 is a compound which is unstable. Even if formed, the compound would act as a very strong oxidizing agent. Why?
Answer:
The oxidation no. of Ag in AgF2 is +2. But, +2 is very unstable oxidation no. of Ag. Hence, when AgF2 is formed, silver accepts an electron and forms Ag+. This decreases the oxidation no. of Ag from +2 to +1. +1 state is more stable. Therefore, AgF2 acts as a very strong oxidizing agent.
11. When there is a reaction between reducing agent and oxidizing agent, a compound is formed which has lower oxidation number if the reducing agent is in excess and a compound is formed which has higher oxidation number if the oxidizing agent is in excess.
Justify the above statement with three examples.
Answer:
When there is a reaction between reducing agent and oxidizing agent, a compound is formed which has lower oxidation number if the reducing agent is in excess and a compound is formed which has higher oxidation number if the oxidizing agent is in excess.
(i) P4 and F2 are reducing and oxidizing agent respectively.
In an excess amount of P4 is reacted with F2, then PF3 would be produced, where the oxidation no. of P is +3.
P4(excess)F2→PF3
If P4 is reacted with excess of F2, then PF5 would be produced, where the oxidation no. of P is +5.
P4+F2(excess)→PF5
(ii) K and O2 acts as a reducing agent and oxidizing agent respectively.
If an excess of K reacts with O2, it produces K2O. Here, the oxidation number of O is -2.
4K(excess)+O2→2K2O−2
If K reacts with an excess of O2, it produces K2O2, where the oxidation number of O is –1.
2K+O2(excess)→K2O−12
(iii) C and O2 acts as a reducing agent and oxidizing agent respectively.
If an excess amount of C is reacted with insufficient amount of O2, then it produces CO, where the oxidation number of C is +2.
C(excess)+O2→CO
If C is burnt in excess amount of O2, then CO2 is produced, where the oxidation number of C is +4.
C+O2(excess)→CO2
12.How do you count for the following observations?
(i) Acidic potassium permanganate and alkaline potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(ii) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:
(i) While manufacturing benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant due to the given reasons.
(a) In a neutral medium, OH– ions are produced in the reaction. Due to that, the cost of adding an acid or a base can be reduced.
(b) KMnO4 and alcohol are homogeneous to each other as they are polar. Alcohol and toluene are homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium compared to heterogeneous medium. Therefore, in alcohol, KMnO4 and toluene can react at a faster rate.
The redox reaction is as given below:
12.1a
(ii) When concentrated H2SO4 is added to an inorganic mixture containing bromide, firstly HBr is produced. HBr, a strong reducing agent, reduces H2SO4 to SO2 with the evolution of bromine’s red vapour.
2NaBr+2H2SO4→2NaHSO4+2HBr
2HBr+H2SO4→Br2+SO2+2H2O
When concentrated H2SO4 I added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, a weak reducing agent, cannot reduce H2SO4 to SO2.
2NaCl+2H2SO4→2NaHSO4+2HCl
13. Find the oxidizing agent, reducing agent, the substance oxidized and reduced for the given reactions:
(i) 2AgBr(s)+C6H6O2(aq)→2Ag(s)+2HBr(aq)+C6H4O2(aq)
(ii) HCHO(l)+2[Ag(NH3)2]+(aq)+3OH–(aq)→2Ag(s)+HCOO–(aq)+4NH3(aq)+2H2O(l)
(iii) HCHO(l)+2Cu2+(aq)+5OH–(aq)→Cu2O(s)+HCOO–(aq)+3H2O(l)
(iv) N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)
(v) Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(aq)+2H2O(l)
Answer:
(i) 2AgBr(s)+C6H6O2(aq)→2Ag(s)+2HBr(aq)+C6H4O2(aq)
C6H6O2 => Oxidized substance
AgBr => Reduced substance
AgBr =>Oxidizing agent
C6H6O2 => Reducing agent
(ii) HCHO(l)+2[Ag(NH3)2]+(aq)+3OH–(aq)→2Ag(s)+HCOO–(aq)+4NH3(aq)+2H2O(l)
HCHO => Oxidized substance
[Ag(NH3)2]+ => Reduced substance
[Ag(NH3)2]+ => Oxidizing agent
HCHO=> Reducing agent
(iii) HCHO(l)+2Cu2+(aq)+5OH–(aq)→Cu2O(s)+HCOO–(aq)+3H2O(l)
HCHO => Oxidized substance
Cu2+ => Reduced substance
Cu2+ => Oxidizing agent
HCHO => Reducing agent
(iv) N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)
N2H4 => Oxidized substance
H2O2 => Reduced substance
H2O2 => Oxidizing agent
N2H4 => Reducing agent
(v) Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(aq)+2H2O(l)
Pb=> Oxidized substance
PbO2 => Reduced substance
PbO2 => Oxidizing agent
Pb => Reducing agent
14. Consider the given reactions:
2S2O2−3(aq)+I2(s)→S4O2−6(aq)+2I–(aq)
S2O2−3(aq)+2Br2(l)+5H2O(l)→2SO2−4(aq)+4Br–(aq)+10H+(aq)
Thiosulphate, the reductant, react differently with bromine and iodine. Why?
Answer:
The average oxidation no. of S inS2O2−3 is +2.
The average oxidation no. of S inS4O2−6 is +2.5.
The oxidation no. of S inS2O2−3 is +2.
The oxidation no. of S inSO2−4 is +6.
As Br2 is a stronger oxidizing agent than I2, it oxidizes S of S2O2−3 to a higher oxidation no. of +6 in SO2−4.
As I2 is a weaker oxidizing agent so it oxidizes S of S2O2−3 ion to a lower oxidation no. that is 2.5 in S4O2−6 ions.
Thus, thiosulphate react differently with I2 and Br2.
15.Among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.Justift this statement with the help of examples
Answer:
F2 can oxidize Cl– to Cl2, Br– to Br2, and I– to I2 as:
F2(aq)+2Cl–(s)→2F–(aq)+Cl2(g)
F2(aq)+2Br–(aq)→2F–(aq)+Br2(l)
F2(aq)+2I–(aq)→2F–(aq)+I2(s)
But, Cl2, Br2, and I2 cannot oxidize F– to F2. The oxidizing power of halogens increases in the order as given below:
I2< Br2< Cl2<F2
Therefore, fluorine is the best oxidant among halogens.
HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Hence, HI and HBr are stronger reductants compared to HCl and HF.
2HI+H2SO4→I2+SO2+2H2O
2HBr+H2SO4→Br2+SO2+2H2O
I– can reduce Cu2+ to Cu+, but Br– cannot.
4I–(aq)+2Cu2+(aq)→Cu2I2(s)+I2(aq)
Therefore, hydrochloric acid is the best reductant among hydrohalic compounds.
Hence, the reducing power of hydrohalic acids increases as given below:
HF< HCl< HBr<HI
16. Why does the given reaction occur?
XeO4−6(aq)+2F–(aq)+6H+(aq)→XeO3(g)+F2(g)+3H2O(l)
What conclusion can be drawn about the compound Na4XeO6 ( of which XeO4−6 is a part) from the reaction?
Answer:
XeO4−6(aq)+2F–(aq)+6H+(aq)→XeO3(g)+F2(g)+3H2O(l)
The oxidation no. of Xe reduces from +8 in XeO4−6 to +6 in XeO3.
The oxidation no. of F increases from -1 in F– to 0 in F2.
Hence, XeO4−6 is reduced on the other hand F– is oxidized. As Na2XeO4−6 (or XeO4−6) is a stronger oxidizing agent compared to F2, this reaction occurs.
17. Consider the following reactions:
(i) H3PO2(aq)+4AgNO3(aq)+2H2O(l)→H3PO4(aq)+4Ag(s)+4HNO3(aq)
(ii) H3PO2(aq)+2CuSO4(aq)+2H2O(l)→H3PO4(aq)+2Cu(s)+H2SO4(aq)
(iii) C6H5CHO(l)+2[Ag(NH3)2]+(aq)+3OH–(aq)→C6H5COO–(aq)+2Ag(s)+4NH3(aq)+2H2O(l)
(iv) C6H5CHO(l)+2Cu2+(aq)+5OH–(aq)→ No change is observed
What can you inference from the reactions about the behavior of Ag+ and Cu2+?
Answer:
Ag+ and Cu2+ behaves as oxidizing agent in reactions (i) and (ii) respectively.
In reaction (iii), Ag+ oxidizes C6H5CHO to C6H5COO–
In reaction (iv), Cu2+ cannot oxidize C6H5CHO.
Therefore, Ag+ is a stronger oxidizing agent compared to Cu2+.
18. Balance the given redox reactions with the help of ion – electron method
(i) MnO–4(aq)+I–(aq)→MnO2(s)+I2(s) (Basic medium)
(ii) MnO–4(aq)+SO2(g)→Mn2+(aq)+H2SO–4 (Acidic medium)
(iii) H2O2(aq)+Fe2+(aq)→Fe3+(aq)+H2O(l) (Acidic medium)
(iv) Cr2−2O7(aq)+SO2(g)→Cr3+(aq)+SO2−(aq) (Acidic medium)
Answer:
(i) MnO–4(aq)+I–(aq)→MnO2(s)+I2(s)
Step 1
The two half reactions are given below:
Oxidation half reaction: I(aq)→I2(s)
Reduction half reaction: MnO–4→MnO2
Step 2
Balance I in oxidation half reaction:
2I–(aq)→I2(s)
Add 2 e– to the right hand side of the reaction to balance the charge:
2I–(aq)→I2(s)+2e–
Step 3
The oxidation no. of Mn has decreased from +7 to +4 in the reduction half reaction. Therefore, 3 electrons are added to the left hand side of the reaction.
MnO–4(aq)+3e–→MnO2(aq)
Add 4 OH– ions to right hand side of the reaction to balance the charge.
MnO–4(aq)+3e–→MnO2(aq)+4OH–
Step 4
There are 6 oxygen atoms on the right hand side and 4 oxygen atoms on the left hand side. Hence, 2 water molecules are added to the left hand side.
MnO–4(aq)+2H2O+3e–→MnO2(aq)+4OH–
Step 5
Equal the no. of electrons on both the sides by multiplying oxidation half reaction by 3 and reduction half reaction by 2:
6I–(aq)→3I2(s)+6e–
2MnO–4(aq)+4H2O+6e–→2MnO2(s)+8OH–(aq)
Step 6
After adding both the half reactions, we get the balanced reaction as given below:
6I–(aq)+2MnO–4(aq)+4H2O(l)→3I2(s)+2MnO2(s)+8OH–(aq)
(ii) MnO–4(aq)+SO2(g)→Mn2+(aq)+H2SO–4
Step 1
Similar to (i), oxidation half reaction is:
SO2(g)+2H2O(l)→HSO–4(aq)+3H+(aq)+2e–(aq)
Step 2
Reduction half reaction is:
MnO–4(aq)+8H+(aq)+5e–→Mn2+(aq)+4H2O(l)
Step 3
Multiply the oxidation half reaction with 5 and the reduction half reaction with 2, then add them. We get the balanced reaction as given below:
2MnO–4(aq)+5SO2(g)+2H2O(l)+H+(aq)→2Mn2+(aq)+5HSO–4(aq)
(iii) H2O2(aq)+Fe2+(aq)→Fe3+(aq)+H2O(l)
Step 1
Similar to (i), oxidation half reaction is:
Fe2+(aq)→Fe3+(aq)+e–
Step 2
Reduction half reaction is:
H2O2(aq)+2H+(aq)+2e–→2H2O(l)
Step 3
Multiply the oxidation half reaction with 2 then add it to the reduction half reaction. We get the balanced reaction as given below:
H2O2(aq)+2Fe2+(aq)2H+(aq)→2Fe3+(aq)2H2O(l)
(iv) Cr2−2O7(aq)+SO2(g)→Cr3+(aq)+SO2−(aq)
Step 1
Similar to (i), oxidation half reaction is:
SO2(g)+2H2O(l)→SO2−4(aq)+4H+(aq)+2e–
Step 2
Reduction half reaction is:
Cr2O2−7(aq)+14H+(aq)+6e–→2Cr3+(aq)+7H2O(l)
Step 3
Multiply the oxidation half reaction with 2 then add it to the reduction half reaction. We get the balanced reaction as given below:
Cr2−2O7(aq)+3SO2(g)+2H+(aq)→2Cr3+(aq)+3SO2−4(aq)+H2O(l)
19. Balance the given redox reactions with the help of ion – electron method and oxidation no. methods. Identify the reducing agent and oxidizing agent.
(i) P4(s)+OH–(aq)→PH3(g)+HPO–2(aq)
(ii) N2H4(l)+ClO–3(aq)→NO(g)+Cl–(g)
(iii) Cl2O7(g)+H2O2(aq)→ClO–2(aq)+O2(g)+H+(aq)
Answer:
(i) The Oxidation no. of P reduces from 0 in P4 to – 3 in PH3
The oxidation no. of P increases from 0 in P4 to + 2 in HPO–2. Therefore, P4 behaves both as areducing agent as well asoxidizing agent in the reaction.
Ion – electron method:
– The oxidation half reaction:
P4(s)→HPO–2(aq)
– Balance atom P:
P4(s)→4HPO–2(aq)
– Add 8 electrons to balance oxidation no.
P4(s)→4HPO–2(aq)+8e–
– Add 12OH– to balance the charge:
P4(s)+12OH−(aq)→4HPO−2(aq)+8e−
– Add 4 H2O to balance H and O atoms:
P4(s)+12OH−(aq)→4HPO–2(aq)+4H2O(l)+8e– ———-(1)
– The reduction half reaction:
P4(s)→PH3(g)
– Balance atom P:
P04(s)→4P−3H3(g)
– Add 12 electrons to balance oxidation no.
P4(s)+12e−→4PH3(g)
– Add 12OH– to balance the charge:
P4(s)+12e−→4PH3(g)+12OH−(aq)
– Add 12 H2O to balance H and O atoms:
P4(s)+12H2O(l)+12e−→4PH3(g)+12OH−(aq) ———- (2)
– Now, multiply the equation (1) by 3 and equation (2) by 2. Then, after adding them, we get the balanced redox reaction as given below:
5P4(s)+12H2O(l)+12HO–(aq)→8PH3(g)+12HPO–2(aq)
(ii)
19.2a
The Oxidation no. of N increases from -2 in N2H4 to –+2 in NO.
The oxidation no. of Cl reduces from +5 in ClO–3 to +-1 in Cl–.
Therefore, N2H4behaves as a reducing agent while ClO–3 behaves as an oxidizing agent in the reaction.
Ion – electron method:
– The oxidation half reaction:
N2H4(l)→NO(g)
– Balance atom N:
N2H4(l)→2NO(g)
– Add 8 electrons to balance oxidation no:
N2H4(l)→2NO(g)+8e−
– Add 8OH– to balance the charge:
N2H4(l)+8OH–(aq)→2NO(g)+8e−
– Add 6 H2O to balance O atoms:
N2H4(l)+8OH−(aq)→2NO(g)+6H2O(l)+8e− ——— (1)
– The reduction half reaction:
ClO–3(aq)→Cl–(aq)
– Add 6 electrons to balance oxidation no.
ClO−3(aq)+6e−→Cl−(aq)
– Add 6OH– ions to balance the charge:
ClO−3(aq)+6e−→Cl−(aq)+6OH−(aq)
– Add 3 H2O to balance O atoms:
ClO−3(aq)+3H2O(l)+6e−→Cl−(aq)+6OH−(aq) ——— (2)
Now, multiply the equation (1) by 3 and equation (2) by 4. Then, after adding them, we get the balanced redox reaction as given below:
3N2H4(l)+4ClO–3(aq)→6NO(g)+4Cl–(aq)+6H2O(l)
Oxidation number method:
– Reduction in the oxidation no. of N = 2 × 4 = 8
– Increment in the oxidation no. of Cl = 1 × 6 = 6
Multiply N2H4 by 3 and ClO–3 by 4 to balance the reduction and increment of the oxidation no. :
3N2H4(l)+4ClO–3(aq)→NO(g)+Cl–(aq)
– Balance Cl and n atoms:
3N2H4(l)+4ClO–3(aq)→6NO(g)+4Cl–(aq)
– Add 6 H2O to balance O atoms:
3N2H4(l)+4ClO–3(aq)→6NO(g)+4Cl–(aq)+6H2O(l)
This is the required reaction equation.
(iii)19.3a
The Oxidation no. of Cldecreases from +7 in Cl2O7to +3 in ClO–2.
The oxidation no. of Oincreases from -1 in H2O2 to 0 in O2.
Therefore, H2O2behaves as a reducing agent while Cl2O7behaves as an oxidizing agent in the reaction.
Ion – electron method:
– The oxidation half reaction:
H2O2(aq)→O2(g)
– Add 2 electrons to balance oxidation no:
H2O2(aq)→O2(g)+2e–
– Add 2OH– to balance the charge:
H2O2(aq)+2OH–(aq)→O2(g)+2e–
– Add 2 H2O to balance O atoms:
H2O2(aq)+2OH–(aq)→O2(g)+2H2O(l)+2e– ——– (1)
– The reduction half reaction:
Cl2O7(g)→ClO–2(aq)
– Balance Cl atoms:
Cl2O7(g)→2ClO–2(aq)
– Add 8 electrons to balance oxidation no.
Cl2O7(g)+8e–→2ClO–2(aq)
– Add 6OH– ions to balance the charge:
Cl2O7(g)+8e–→2ClO–2(aq)+6OH–(aq)
– Add 3 H2O to balance O atoms:
Cl2O7(g)+3H2O(l)+8e–→2ClO–2(aq)+6OH–(aq)
Now, multiply the equation (1) by 4. Then, adding equation (1) and (2), we get the balanced redox reaction as given below:
Cl2O7(g)+4H2O2(aq)+2OH–(aq)→2ClO–2(aq)+4O2(g)+5H2O(l)
Oxidation number method:
– Reduction in the oxidation no. of Cl2O7 = 4× 2 = 8
– Increment in the oxidation no. of H2O2 = 2× 1 = 2
Multiply H2O2 by 4 and O2 by 4 to balance the reduction and increment of the oxidation no. :
3N2H4(l)+4ClO–3(aq)→NO(g)+Cl–(aq)
– Balance Cl and n atoms:
Cl2O7(g)+4H2O2(aq)→2ClO–2(aq)+4O2(g)
– Add 3 H2O to balance O atoms:
Cl2O7(g)+4H2O2(aq)→2ClO–2(aq)+4O2(g)+3H2O(l)
– Add 2OH– and 2H2O to balance H atoms:
Cl2O7(g)+4H2O2(aq)2OH–(aq)→2ClO–2(aq)+4O2(g)+5H2O(l)
This is the required reaction equation.
20.What information can be drawn from the reactions given?
(CN)2(g)+2OH–(aq)→CN–(aq)+CNO–(aq)+H2O(l)
Answer:
The oxidation no. of C in (CN)2, CN– and CNO– are +3, +2 and +4 respectively.
Let the oxidation no. of C be y.
(CN)2
2(y – 3) = 0
Therefore, y = 3
CN–
y – 3 = -1
Therefore, y = 2
CNO–
y – 3 -2 = -1
Therefore, y = 4
The oxidation no. of C in the reaction is:
Oxidation no. of C in (CN)2 is +3
Oxidation no. of C in CN– is +2
Oxidation no. of C in CNO– is +4
We can see that the same compound is oxidized and reduced simultaneously in the reaction.
The reactions in which the same compound is oxidized and reduced is known as disproportionation reaction. Then, we can say that alkaline decomposition of cyanogens is a disproportionation reaction.
21. Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.
Answer:
The reaction is as given below:
Mn3+(aq)→Mn2+(aq)+MnO2(s)+H+(aq)
The oxidation half reaction:
Mn3+(aq)→MnO2(s)
Add 1 electron to balance the oxidation no. :
Mn3+(aq)→MnO2(s)+e–
Add 4H+ ions to balance the charge:
Mn3+(aq)→MnO2(s)+e–+4H+(aq)
Add 2 H2O to balance O atoms and H+ ions:
Mn3+(aq)+2H2O(l)→MnO2(s)+e–+4H+(aq) ———— (1)
The reduction half reaction:
Mn3+(aq)→Mn2+(aq)
Add 1 electron to balance the oxidation no. :
Mn3+(aq)+e–→Mn2+(aq) ——— (2)
Add equation (1) and (2) to get the balanced chemical equation:
2Mn3+(aq)+2H2O(l)→MnO2(s)+2Mn2+(aq)+4H+(aq)
22.Consider the elements:
Cs, I, Ne and F
(i) Which element exhibit only negative oxidation no.?
(ii) Which element exhibit only positive oxidation no.?
(iii) Which element exhibit both negative and positive oxidation no.?
(iv) Which element exhibits neither negative nor positive oxidation no.?
Answer:
(i) F exhibits only negative oxidation no. That is -1.
(ii) Cs exhibits only positive oxidation no. That is +1.
(iii) I exhibits both negative and positive oxidation no. That is -1, +1, +3, +5 and +7.
(iv) Ne exhibits neither negative nor positive oxidation no. That is 0.
23. It is said that chlorine is harmful in excess. To purify water, chlorine is used. Hence, it becomes necessary to remove excess of chlorine from water for which, water is treated with sulphur dioxide. Obtain a balanced equation for this redox reaction:
Answer:
The redox reaction is as given below:
Cl2(s)+SO2(aq)+H2O(l)→Cl–(aq)+SO2−4(aq)
The oxidation half reaction:
SO2(aq)→SO2−4(aq)
Add 2 electrons to balance the oxidation no. :
SO2(aq)→SO2−4(aq)+2e–
Add 4H+ ions to balance the charge:
SO2(aq)→SO2−4(aq)+4H+(aq)+2e–
Add 2 H2O to balance O atoms and H+ ions:
SO2(aq)+2H2O→SO2−4(aq)+4H+(aq)+2e– ——— (1)
The reduction half reaction:
Cl2(s)→Cl−(aq)
Balance Cl atoms:
Cl2(s)→2Cl–(aq)
Add 2 electrons to balance the oxidation no. :
Cl2(s)+2e–→2Cl–(aq) ——— (2)
Add equation (1) and (2) to get the balanced chemical equation:
Cl2(s)+SO2(aq)+2H2O(l)→2Cl–(aq)+SO2−4(aq)+4H+(aq)
24. Answer the given questions referring to the periodic table:
(i) Which non – metals can show disproportionation reaction?
(ii) Which three metals shows disproportionation reaction?
Answer:
One of the reacting elements always has an element that can exist in at least 3 oxidation numbers.
(i) The non – metals which can show disproportionation reactions are P, Cl and S.
(ii) The three metals which can show disproportionation reactions are Mn, Ga and Cu.
25. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?
Answer:
The balanced reaction is as given below:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
4NH3 = 4 × 17 g = 68 g
5O2 = 5 × 32 g = 160 g
4NO = 4 × 30 g = 120 g
6H2O = 6 × 18 g = 108 g
Thus, NH3 (68 g) reacts with O2 ( 20 g )
Therefore, 10 g of NH3 reacts with 160×1068 g = 23.53 g of O2
But only 20 g of O2 is available.
Hence, O2 is a limiting reagent.
Now, 160 g of O2 gives 120×20160 g of N = 15 g of NO.
Therefore, max of 15 g of nitric oxide can be obtained.
26. Predict whether the reaction is feasible or not between the given elements.
(i) Fe3+(aq) and I–(aq)
(ii)Ag+(aq) and Cu(s)
(iii) Fe3+(aq)and Cu(s)
(iv) Ag(s) and Fe3+(aq)
(v) Br2(aq) and Fe2+(aq)
Answer:
(i) Fe3+(aq) and I–(aq)
2Fe3+(aq)+2I–(aq)→2Fe2+(aq)+I2(s)
Oxidation half reaction: 2I–(aq)→I2(s)+2e– ; E∘=−0.54V
Reduction half reaction: [Fe3+(aq)+e–→Fe2+(aq)]×2; ; E∘=+0.77V
2Fe3+(aq)+2I–→2Fe2+(aq)+I2(s);; E∘=+0.23V
E∘for the overall reaction is positive. Therefore, the reaction between Fe3+(aq) and I–(aq) is feasible.
(ii) Ag+(aq) and Cu(s)
2Ag+(aq)+Cu(s)→2Ag(s)+Cu2+(aq)
Oxidation half reaction: Cu(s)→Cu2+(aq)+2e–; E∘=−0.34V
Reduction half reaction: [Ag+(aq)+e–→Ag(s)]×2; E∘=+0.80V
2Ag+(aq)+Cu(s)→2Ag(s)+Cu2+; E∘=+0.46V
E∘ for the overall reaction is positive. Therefore, the reaction between Ag+(aq) and Cu(s) is feasible.
(iii) Fe3+(aq)and Cu(s)
2Fe3+(aq)+Cu(s)→2Fe2+(s)+Cu2+(aq)
Oxidation half reaction: Cu(s)→Cu2+(aq)+2e–; E∘=−0.34V
Reduction half reaction: [Fe3+(aq)+e–→Fe2+(s)]×2; E∘=+0.77V
2Fe3+(aq)+Cu(s)→2Fe2+(s)+Cu2+(aq); E∘=+0.43V
E∘ for the overall reaction is positive. Therefore, the reaction between Fe3+(aq)and Cu(s) is feasible.
(iv) Ag(s) and Fe3+(aq)
Ag(s)+2Fe3+(aq)→Ag+(aq)+Fe2+(aq)
Oxidation half reaction: Ag(s)→Ag+(aq)+e–; E∘=−0.80V
Reduction half reaction: Fe3+(aq)+e–→Fe2+(aq); E∘=+0.77V
Ag(s)+Fe3+(aq)→Ag+(aq)+Fe2+(aq); E∘=−0.03V
E∘ for the overall reaction is positive. Therefore, the reaction between Ag(s) and Fe3+(aq) is feasible.
(v) Br2(aq) and Fe2+(aq)
Br2(s)+2Fe2+(aq)→2Br–(aq)+2Fe3+(aq)
Oxidation half reaction: [Fe2+(aq)→Fe3+(aq)+e–]×2; E∘=−0.77V
Reduction half reaction: Br2(aq)+2e–→2Br–(aq); E∘=+1.09V
Br2(s)+2Fe2+(aq)→2Br–(aq)+2Fe3+(aq); E∘=−0.32V
E∘ for the overall reaction is positive. Therefore, the reaction between Br2(aq) and Fe2+(aq) is feasible.
27. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes
(ii) An aqueous solution AgNO3 with platinum electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Answer:
(i) AgNO3 ionizes in aqueous solution to form Ag+ and NO−3 ions.
On electrolysis, either Ag+ ion or H2O molecule can be decreased at cathode. But the reduction potential of Ag+ ions is higher than that of H2O.
Ag+(aq)+e−→Ag(s); E∘=+0.80V
2H2O(l)+2e−→H2(g)+2OH−(aq); E∘=−0.83V
Therefore, Ag+ ions are decreased at cathode. Same way, Ag metal or H2O molecules can be oxidized at anode. But the oxidation potential of Ag is greater than that of H2O molecules.
Ag(s)→Ag+(aq)+e−; E∘=−0.80V
2H2O(l)→O2(g)+4H+(aq)+4e−; E∘=−1.23V
Hence, Ag metal gets oxidized at anode.
(ii) Pt cannot be oxidized very easily. Therefore, at anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are decreased and get deposited.
(iii) H2SO4 ionizes in aqueous solutions to give H+ and SO2−4 ions.
H2SO4(aq)→2H+(aq)+SO2−4(aq)
On electrolysis, either of H2O molecules or H+ ions can get decreased at cathode. But the decreased potential of H+ ions is higher than that of H2O molecules.
2H+(aq)+2e−→H2(g); E∘=0.0V
2H2O(aq)+2e−→H2(g)+2OH−(aq); E∘=−0.83V
Therefore, at cathode, H+ ions are decreased to free H2 gas.
On the other hand, at anode, either of H2O molecules or SO2−4 ions can be oxidized. But the oxidation of SO2−4 involves breaking of more bonds than that of H2O molecules. Therefore, SO2−4 ions have lower oxidation potential than H2O. Hence, H2O is oxidized at anode to free O2 molecules.
(iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl− ions as:
CuCl2(aq)→Cu2+(aq)+2Cl−(aq)
CuCl2(aq)→Cu2+(aq)+2Cl−(aq)
On electrolysis, either of Cu2+ ions or H2O molecules can get decreased at cathode. But the decreased potential ofCu2+ is more than that of H2O molecules.
Cu2+(aq)+2e−→Cu(aq); E∘=+0.34V
H2O(l)+2e−→H2(g)+2OH−; E∘=−0.83V
Therefore, Cu2+ ions are decreased at cathode and get deposited. In the same way, at anode, either of Cl− orH2O is oxidized. The oxidation potential of H2O is higher than that of Cl−.
2Cl−(aq)→Cl2(g)+2e−; E∘=+0.34V
2H2O(l)→O2(g)+4H+(aq)+4e−; E∘=−1.23V
But oxidation of H2O molecules occurs at a lower electrode potential compared to that of Cl− ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl− ions are oxidized at the anode to liberate Cl2 gas.
28. Arrange the given metals in the order in which they displace each other from the solution of their salts.
(i) Al
(ii) Fe
(iii) Cu
(iv) Zn
(v) Mg
Answer:
A metal with stronger reducing power displaces another metal with weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is as given below:
Cu < Fe < Zn < Al < Mg
Therefore, Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu
29.The standard electrode potentials are given of the following elements:
K+/K = –2.93V
Ag+/Ag = 0.80V
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37V
Cr3+/Cr = –0.74V
Arrange these metals in their increasing order of reducing power.
Answer:
The reducing agent is stronger as the electrode potential decreases. Hence, the increasing order of the reducing power of the given metals is as given below:
Ag < Hg < Cr < Mg < K
30. Depict the galvanic cell in which the reaction is:
Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s)
Show the following:
(i) Which of the electrode is negatively charged?
(ii) Name the carriers of the current in the cell.
(iii) Write the individual reaction at each electrode.
Answer:
The galvanic cell corresponding to the given redox reaction can be shown as:
Zn|Zn2+(aq)||Ag+(aq)|Ag
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the leaving electrons accumulate on this electrode.
(ii) The carriers of current are ions in the cell.
(iii) Reaction at Zn electrode is shown as:
Zn(s)→Zn2+(aq)+2e−
Reaction at Ag electrode is shown as:
Ag+(aq)+e−→Ag(s)
0 Comments