Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 6 Redox Reactions Textbook Exercise Questions and Answers.
Redox Reactions Class 11 Exercise Question Answers Solutions Maharashtra Board
Class 11 Chemistry Chapter 6 Exercise Solutions Maharashtra Board
Chemistry Class 11 Chapter 6 Exercise Solutions
1. Choose the most correct option
Question A.
Oxidction numbers of Cl atoms marked as Cl
a
and Cl
b
in CaOCl
2
(bleaching powder) are
a. zero in each
b. -1 in Cl
a
and +1 in Cl
b
c. +1 in Cl
a
and -1 in Cl
b
d. 1 in each
Answer:
b. -1 in Cl
a
and +1 in Cl
b
Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H
2
→ Cu + H
2
O
b. Fe
2
O
3
+ 3CO
2
→ 2Fe + 3CO
2
c. 2K + F
2
→ 2KF
d. BaCl
2
+ H
2
SO
4
→ BaSO
4
+ 2HCl
Answer:
d. BaCl
2
+ H
2
SO
4
→ BaSO
4
+ 2HCl
Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A
2
(BC
3
)
2
b. A
3
(BC
4
)
2
c. A
3
(B
4
C)
2
d. ABC
2
Answer:
b. A
3
(BC
4
)
2
Question D.
The coefficients p, q, r, s in the reaction
\(\mathrm{pCr}_{2} \mathrm{O}_{7}^{2-}\) + q Fe
2⊕
→ r Cr
3⊕
+ s Fe
3⊕
+ H
2
O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6
Question E.
For the following redox reactions, find the correct statement.
Sn
2⊕
+ 2Fe
3⊕
→ Sn
4⊕
+ 2Fe
2⊕
a. Sn
2⊕
is undergoing oxidation
b. Fe
3⊕
is undergoing oxidation
c. It is not a redox reaction
d. Both Sn
2⊕
and Fe
3⊕
are oxidised
Answer:
a. Sn
2⊕
is undergoing oxidation
Question F.
Oxidation number of carbon in H
2
CO
3
is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4
Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O
2
b. Mn(II)O
2
c. Mn(III)O
2
d. Mn(IV)O
2
Answer:
d. Mn(IV)O
2
Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –\(\frac {1}{2}\)
d. 0
Answer:
c. –\(\frac {1}{2}\)
Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F
Question K.
The process SO
2
→ S
2
Cl
2
is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction
2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl
2
ii. Tl
2
SO
4
iii. SnO
2
iv. Cr
2
O
3
3. Answer the following questions
Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH
4
+ 2O
2
→ CO
2
+ 2H
2
O
In CH
4
, the oxidation state of carbon is -4 while in CO
2
, the oxidation state of carbon is +4.
Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:
C. Calculate the oxidation number of underlined atoms.
a. H
2
S
O
4
b. H
N
O
3
c. H
3
P
O
3
d. K
2
C
2
O
4
e. H
2
S
4
O
6
f.
Cr
2
O
7
2-
g. NaH
2
P
O
4
Answer:
i. H
2
S
O
4
Oxidation number of H = +1
Oxidation number of O = -2
H
2
SO
4
is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H
2
SO
4
= 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H
2
SO
4
= +6
ii. H
N
O
3
Oxidation number of H = +1
Oxidation number of O = -2
HNO
3
is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO
3
= 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO
3
= +5
iii. H
3
P
O
3
Oxidation number of O = -2
Oxidation number of H = +1
H
3
PO
3
is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H
3
PO
3
= +3
iv. K
2
C
2
O
4
Oxidation number of K = +1
Oxidation number of O = -2
K
2
C
2
O
4
is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +\(\frac {6}{2}\)
∴ Oxidation number of C in K
2
C
2
O
4
= +3
v. H
2
S
4
O
6
Oxidation number of H = +1
Oxidation number of O = -2
H
2
S
4
O
6
is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +\(\frac {10}{4}\)
∴ Oxidation number of S in H
2
S
4
O
6
= +2.5
vi.
Cr
2
O
7
2-
Oxidation of O = -2
Cr
2
O
7
2-
is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +\(\frac {12}{2}\)
∴ Oxidation number of Cr in Cr
2
O
7
2-
= +6
vii. NaH
2
P
O
4
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH
2
PO
4
is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH
2
PO
4
= +5
Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu
2
O
(s)
+ Cu
2
S
(s)
→ 6Cu
(s)
+ SO
2(g)
b. HF
(aq)
+ OH
–
(aq)
→ H
2
O
(l)
+ F
–
(aq)
c. I
2(aq)
+ 2 S
2
O
3
2-
(aq)
→ S
4
O
6
2-
(aq)
+ 2I
–
(aq)
Answer:
i. 2Cu
2
O
(s)
+ Cu
2
S
(s)
→ 6Cu
(s)
+ SO
2(g)
a. Write oxidation number of all the atoms of reactants and products.
b. Identify the species that undergoes change in oxidation number.
c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.
Result:
- The given reaction is a redox reaction.
- Oxidant/oxidising agents (Reduced species): Cu 2 O/ Cu 2 S
- Reductant/reducing agent (Oxidised species): Cu 2 S
[Note: Cu in both Cu 2 O and Cu 2 S undergoes reduction. Hence, both Cu 2 O and Cu 2 S can be termed as oxidising agents in the given reaction.]
ii. HF
(aq)
+ OH
–
(aq)
→ H
2
O
(l)
+ F
–
(aq)
a. Write oxidation number of all the atoms of reactants and products.
b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.
iii. I
2(aq)
+ 2 S
2
O
3
2-
(aq)
→ S
4
O
6
2-
(aq)
+ 2I
–
(aq)
a. Write oxidation number of all the atoms of reactants and products.
b. Identify the species that undergoes change in oxidation number.
c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.
Result:
- The given reaction is a redox reaction.
- Oxidant/oxidising agent (Reduced species): I 2
- Reductant/reducing agent (Oxidised species): S 2 O 3 2-
Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg
2+
b. Cu / Cu
2+
c. O
2
/ O
2-
d. Cl
2
/ Cl
–
Answer:
a. Mg / Mg
2+
Here, Mg loses two electrons to form Mg
2+
ion.
\(\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Mg / Mg
2+
is an oxidized state.
b. Cu/Cu
2+
Here, Cu loses two electrons to form Cu
2+
ion.
\(\mathrm{Cu}_{(\mathrm{s})} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Cu/Cu
2+
is in an oxidized state.
c. O
2
/ O
2-
Here, each O gains two electrons to form O
2-
ion.
\(\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}_{(\mathrm{aq})}^{2-}\)
Hence, O
2
/ O
2-
is in a reduced state.
d. Cl
2
/ Cl
–
Here, each Cl gains one electron to form Cl
–
ion.
\(\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence, Cl
2
/ Cl
–
is in a reduced state.
Question F.
Justify the following reaction as redox reaction.
2 Na
2(s)
+ S
(s)
→ Na
2
S
(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na
(s)
+ S
(s)
→ 2Na
+
+ S
2-
ii. Charge development suggests that each sodium atom loses one electron to form Na
+
and sulphur atom gains two electrons to form S
2-
. This can be represented as follows:
iii. When Na is oxidised to Na
2
S, the neutral Na atom loses electrons to form Na
+
in Na
2
S while the elemental sulphur gains electrons and forms S
2-
in Na
2
S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.
Question G.
Provide the stock notation for the following compounds : HAuCl
4
, Tl
2
O, FeO, Fe
2
O
3
, MnO and CuO.
Answer:
Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)
4
–
b. Na
2
S
2
O
3
c. H
3
BO
3
Answer:
i. Cr(OH)
4
–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)
4
–
is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)
4
–
= +3
ii. Na
2
S
2
O
3
Oxidation number of Na = +1
Oxidation number of O = -2
Na
2
S
2
O
3
is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +\(\frac {4}{2}\)
∴ Oxidation number of S in Na
2
S
2
O
3
= +2
iii. H
3
BO
3
Oxidation number of H = +1
Oxidation number of O = -2
H
3
BO
3
is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H
3
BO
3
= +3
Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl
2
(E
0
= 1.36 V) and Br
2
(E
0
= 1.09 V)
b. \(\mathrm{MnO}_{4}^{\Theta}\) (E0 = 1.51 V) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\) (E
0
= 1.33 V)
Answer:
a. Cl
2
has a larger positive value of E
0
than Br
2
. Thus, Cl
2
is a stronger oxidizing agent than Br
2
.
b. \(\mathrm{MnO}_{4}^{\Theta}\) has larger positive value of E
0
than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\). Thus, \(\mathrm{MnO}_{4}^{\Theta}\) is stronger oxidizing agent than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\)
Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E
0
= – 3.05 V) and Mg(E
0
= – 2.36 V)
b. Zn(E
0
= – 0.76 V) and Fe(E
0
= – 0.44 V)
Answer:
a. Li has a larger negative value of E
0
than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E
0
than Fe. Thus, Zn is a stronger reducing agent than Fe.
4. Balance the reactions/equations :
Question A.
Balance the following reactions by oxidation number method
Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \quad(\text { acidic })\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(a q)}^{2-}+\mathrm{SO}_{3(a)}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H
2
O to the right-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:
ii. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{(a q)} \quad \text { (basic) }\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{ }_{(\mathrm{aq})}\)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 2 atoms of Mn.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}\)
Step 3: Balance ‘O’ atoms by adding H
2
O to the right-hand side.
\(2 \mathrm{MnO}_{4(a q)}^{-}+\mathrm{Br}_{(2 q)}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3 \text { (aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H
+
on the left-hand side.
iii. H
2
SO
4(aq)
+ C
(s)
→ CO
2(g)
+ SO
2(g)
+ H
2
O
(l)
(acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H
2
SO
4(aq)
+ C
(s)
→ CO
2(g)
+ SO
2(g)
+ H
2
O
(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 2 atoms of S.
2H
2
SO
4(aq)
+ C
(s)
→ CO
2(g)
+ 2SO
2(g)
+ H
2
O
(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H
2
SO
4(aq)
+ C
(s)
→ CO
2(g)
+ 2SO
2(g)
+ H
2
O
(l)
+ H
2
O
(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H
2
SO
4(aq)
+ C
(s)
→ CO
2
+ 2SO
2(g)
+ H
2
O
(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H
2
SO
4(aq)
+ C
(s)
→ CO
2(g)
+ 2SO
2(g)
+ H
2
O
(l)
iv. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.
To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.
Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.
Question B.
Balance the following redox equation by half reaction method
Answer:
i. H
2
C
2
O
4(aq)
+ \(\mathrm{MnO}_{4(a q)}^{-}\) → CO
2(g)
+ \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H
2
O to the right side of reduction half equation.
Step 3: Balance H atoms by adding H
+
ions to the side with less H. Hence, add 2H
+
ions to the right side of oxidation half equation and 8H
+
ions to the left side of reduction half equation.
Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.
Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.
ii. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{SnO}_{2(\mathrm{aq})}^{2-} \longrightarrow \mathrm{SnO}_{3(\mathrm{aq})}^{2-}+\mathrm{Bi}_{(\mathrm{s})}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
Step 2: Balance half equations for O atoms by adding H
2
O to the side with less O atoms. Add 1H
2
O to left side of oxidation half equation and 3H
2
O to the right side of reduction half equation.
Step 3: Balance H atoms by adding H
+
ions to the side with less H. Hence, add 2H
+
ions to the right side of oxidation half equation and 3H
+
ions to the left side of reduction half equation.
Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.
Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.
Reaction occurs in basic medium. However, H
+
ions cancel out and the reaction is balanced. Hence, no need to add OH
–
ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:
5. Complete the following table :
Assign oxidation number to the underlined species and write Stock notation of compound
| Compound | Oxidation number | Stock notation |
| Au Cl 3 | …………….. | …………….. |
| Sn Cl 2 | …………….. | …………….. |
| \(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\) | …………….. | …………….. |
| \(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\) | …………….. | …………….. |
| H 3 As O 3 | …………….. | …………….. |
Answer:
| Compound | Oxidation number | Stock notation |
| Au Cl 3 | +3 | Au(III)Cl 3 |
| Sn Cl 2 | +2 | Sn(II)Cl 2 |
| \(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\) | +5 | V 2 (V)\(\mathrm{O}_{7}^{4-}\) |
| \(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\) | +4 | Pt(IV)\(\mathrm{Cl}_{6}^{2-}\) |
| H 3 As O 3 | +3 | H 3 As(III)O 3 |
11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers
Can you tell? (Textbook Page No. 81)
Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.
Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.
Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.
Can you recall? (Textbook Page No. 81)
Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.
Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH
4
+ 2O
2
→ CO
2
+ 2H
2
O + Heat + Light
Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.
Try this. (Textbook Page No. 82)
Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.
| Reactants | Products |
| Zn (s) + ———— (aq) | ————- (aq) + Cu (s) |
| Cu (s) + 2Ag + (aq) | —————– + ————– |
| ———– + ————- | \( \mathrm{Co}_{(\mathrm{aq})}^{2+}\) + Ni (s) |
Answer:
Try this (Textbook Page No. 88)
Question 1.
Classify the following unbalanced half equations as oxidation and reduction.
Answer:
