Answer the following questions.

Question 14 Marks

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_2O_{(s)} + Cu_2S_{(s)} \rightarrow 6Cu_{(s)} + SO_{2(g)}$
b. $ HF_{(aq)} + OH^–_{(aq)} \rightarrow H_2O_{(l)} + F^–_{(aq)}$
c. $I_{2(aq)} + {2 S_2O_3^{2-}}_{(aq)} \rightarrow {S_4O_6^{2-}}_{(aq)} + 2I^–_{(aq)}$

Answer

i. $2Cu_2O_{(s)} + Cu_2S_{(s)} \rightarrow 6Cu_{(s)} + SO_{2(g)}$_{
}
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$): Cu_2O/ Cu_2S$
Reductant/reducing agent $($Oxidised species$): Cu_2S$

[Note: $Cu$ in both $Cu_2O$ and $Cu_2S$ undergoes reduction. Hence, both $Cu_2O$ and $Cu_2S$ can be termed as oxidising agents in the given reaction.]
ii. $HF_{(aq)} + OH^–_{(aq)} \rightarrow H_2O_{(l)} + F^–_{(aq)}$_
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. $I_{2(aq)} + {2 S_2O_3^{2-}}_{(aq)} \rightarrow {S_4O_6^{2-}}_{(aq)} + 2I^–_{(aq)}$
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_2O_3^{2-}$

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Question 24 Marks

Calculate the oxidation number of underlined atoms.
a. $H_2SO_4$
b. $HNO_3$
c. $H_3PO_3$
d. $K_2C_2O_4$
e. $H_2S_4O_6$
f. $Cr_2O_7^{2-}$
g. $NaH_2PO_4$

Answer

i. $H _2 SO _4$
Oxidation number of $H =+1$
Oxidation number of $O =-2$
$H _2 SO _4$ is a neutral molecule.
$\therefore$ Sum of the oxidation numbers of all atoms of $H _2 SO _4=0$
$\therefore 2 \times($ Oxidation number of $H )+($ Oxidation number of $S )+4 \times($ Oxidation number of $O )$ $=0$
$\therefore 2 \times(+1)+$ (Oxidation number of $S)+4 \times(-2)=0$
$\therefore$ Oxidation number of $S+2-8=0$
$\therefore$ Oxidation number of $S$ in $H _2 SO _4=+6$ii. $HNO _3$
Oxidation number of $H =+1$
Oxidation number of $O =-2$
$HNO _3$ is a neutral molecule.
$\therefore$ Sum of the oxidation numbers of all atoms of $HNO _3=0$
$\therefore$ (Oxidation number of $H )+($ Oxidation number of $N )+3 \times($ Oxidation number of $O )=0$
$\therefore(+1)+$ (Oxidation number of $N)+3 \times(-2)=0$
$\therefore$ Oxidation number of $N +1-6=0$
$\therefore$ Oxidation number of $N$ in $HNO _3=+5$
iii. $H _3 PO _3$
Oxidation number of $O =-2$
Oxidation number of $H =+1$
$H _3 PO _3$ is a neutral molecule.
$\therefore$ Sum of the oxidation numbers of all atoms $=0$
$\therefore 3 \times($ Oxidation number of $H )+$ (Oxidation number of $P )+3 \times($ Oxidation number of $O )$ $=0$
$\therefore 3 \times(+1)+$ (Oxidation number of $P)+3 \times(-2)=0$
$\therefore$ 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.
$\therefore$ Sum of the oxidation number of all atoms $=0$
$\therefore 2 \times$ (Oxidation number of $K )+2 \times$ (Oxidation number of $C )+4 \times$ (Oxidation number of
O) $=0$
$\therefore 2 \times(+1)+2 \times($ Oxidation number of $C)+4 \times(-2)=0$
$\therefore 2 \times$ (Oxidation number of C) $+2-8=0$
$\therefore 2 \times$ (Oxidation number of $C )=+6$
$\therefore$ Oxidation number of $C =+\frac{6}{2}$
$\therefore$ 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.
$\therefore$ Sum of the oxidation numbers of all atoms $=0$
$\therefore 2 \times($ Oxidation number of $H )+4 \times$ (Oxidation number of $S )+6 \times$ (Oxidation number of $O )=0$
$\therefore 2 \times(+1)+4 \times($ Oxidation number of S) $+6 \times(-2)=0$
$\therefore 4 \times$ (Oxidation number of S) $+2-12=0$
$\therefore 4 \times($ Oxidation number of $S)=+10$
$\therefore$ Oxidation number of $S =+\frac{10}{4}$
$\therefore$ 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.
$\therefore$ Sum of the oxidation numbers of all atoms $=-2$
$\therefore 2 \times$ (Oxidation number of $Cr )+7 \times$ (Oxidation number of $O )=-2$
$\therefore 2 \times$ (Oxidation number of $Cr )+7 \times(-2)=-2$
$\therefore 2 \times$ (Oxidation number of $Cr )-14=-2$
$\therefore 2 \times$ (Oxidation number of $Cr )=-2+14$
$\therefore$ Oxidation number of $Cr =+\frac{12}{2}$
$\therefore$ Oxidation number of $Cr$ in $Cr _2 O _7^{2-}=+6$
vii. $NaH _2 PO _4$
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 $Na )+2 \times$ (Oxidation number of $H )+$ (Oxidation number of $P )+4 \times$
(Oxidation number of $O )=0$
$(+1)+2 \times(+1)+($ Oxidation number of $P)+4 \times(-2)=0$
(Oxidation number of $P$ ) $+3-8=0$
Oxidation number of $P$ in $NaH _2 PO _4=+5$

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Question 34 Marks

Balance the following redox equation by half reaction method

Answer

Balance the following redox equation by half reaction method

i. $H _2 C _2 O _{4( aq )}+ MnO _{4(a q)}^{-} \rightarrow CO _{2( g )}+ Mn _{( aq )}^{2+}$
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 $4H_2O$ 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 $2 H ^{+}$ions to the right side of oxidation half equation and $8 H ^{+}$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. $Bi ( OH )_{3( s )}+ SnO _{2( aq )}^{2-} \longrightarrow SnO _{3( aq )}^{2-}+ Bi _{( s )}$
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 $H_2O$ to the side with less O atoms. Add $1H_2O$ to left side of oxidation half equation and $3H_2O$ 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:

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Question 44 Marks

Balance the following reactions by oxidation number method

Answer

i. $Cr _2 O _{7( aq )}^{2-}+ SO _{3( aq )}^{2-} \longrightarrow Cr _{( aq )}^{3+}+ SO _{4( aq )}^{2-} \quad$ ( acidic )
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$
Cr _2 O _{7(a q)}^{2-}+ SO _{3(a)}^{2-} \longrightarrow 2 Cr _{( aq )}^{3+}+ SO _{4( 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 $4 H _2 O$ to the right-hand side.
$
Cr _2 O _{7( aq )}^{2-}+3 SO _{3( aq )}^{2-} \longrightarrow 2 Cr _{( aq )}^{3+}+3 SO _{4( aq )}^{2-}+4 H _2 O _{(l)}
$
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add $8 H$ on the left-hand side.
$
Cr _2 O _{7( aq )}^{2-}+3 SO _{3( aq )}^{2-}+8 H _{( aq )}^{+} \longrightarrow 2 Cr _{( aq )}^{3+}+3 SO _{4( aq )}^{2-}+4 H _2 O _{(l)}
$
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. $MnO _{4( aq )}^{-}+ Br _{( aq )}^{-} \longrightarrow MnO _{2( s )}+ BrO _3^{-}(a q) \quad$ (basic)
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$MnO _{4( aq )}^{-}+ Br _{( aq )}^{-} \longrightarrow MnO _{2( s )}+ BrO _{3( aq )}^{-}$
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 $Mn$.
$2 MnO _{4( aq )}^{-}+ Br _{( aq )}^{-} \longrightarrow 2 MnO _{2( s )}+ BrO _{3( aq )}^{-}$
Step 3: Balance ' $O ^{\text {at }}$ atoms by adding $H _2 O$ to the right-hand side.
$2 MnO _{4(a q)}^{-}+ Br _{(2 q)}^{-} \longrightarrow 2 MnO _{2( s )}+ BrO _{3( aq )}^{-}+ H _2 O _{(l)}$
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add $2 H ^{+}$on the left-hand side.

iii. $H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+ SO _{2( g )}+ H _2 O _{( l )}$ (acidic)
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$
H _2 SO _{4( aq )}+ C _{( s )} \rightarrow 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 $S$.
$
2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+2 SO _{2( g )}+ H _2 O _{( l )}
$
Step 3: Balance ' $O ^{\prime}$ atoms by adding $H 2 O$ to the right-hand side.
$
2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+2 SO _{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.
$
2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _2+2 SO _{2( g )}+ H _2 O _{( l )}
$
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: $2 H _2 SO _{4( aq )}+ C _{( s )} \rightarrow CO _{2( g )}+2 SO _{2( g )}+ H _2 O _{( l )}$
iv. $Bi ( OH )_{3( s )}+ Sn ( OH )_{3( aq )}^{-} \longrightarrow Bi _{( s )}+ Sn ( OH )_{6( aq )}^{2-}$ (basic)
Step 1: Write skeletal equation and balance the elements other than $O$ and $H$.
$
Bi ( OH )_{3( s )}+ Sn ( OH )_{3( aq )}^{-} \longrightarrow Bi _{( s )}+ Sn ( OH )_{6( 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.

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Question 54 Marks

Explain working of Daniel cell.

Answer

$i.$ When a circuit is complete, the zinc atoms on zinc plates spontaneously lose electrons which are picked up in the external circuit.
$ii.$ The electrons flow from the zinc plate to copper plate through wire.
$iii. \ Cu^{2+}$ ions in the second container receive these electrons through the copper plate and are reduced to copper atoms which get deposited on the copper plate.
$iv.$ Here, zinc plate acts as anode (negative electrode) and the copper plate acts as cathode $($positive electrode$).$
Oxidation$: Zn _{(s)} \longrightarrow Zn _{(\text {aq })}^{2+}+2 e$
Reduction$: Cu _{\left( x _1\right)}^{2+}+2 e ^{-} \longrightarrow Cu _{( s )}$
$v.$ Thus, when two half reactions, namely, oxidation and reduction, are allowed to take place in separate containers and provision is made for completing the electrical circuit, electron transfer take place through the circuit.
$vi.$ This results in flow of electric current in the circuit as indicated by deflection in voltmeter.
$vii.$ Thus, in Daniel cell, electricity is generated by redox reaction.

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Question 64 Marks

Explain construction of Daniel cell.

Answer

The zinc and copper plates are connected by an electric wire through a switch and voltmeter.
The solution in two containers are connected by salt bridge $(U-$shaped glass tube containing a gel of $KCl \ or \ NH_4NO_3$ in agar$-$agar$).$
When switch is on, electrical circuit is complete as indicated by the deflection in the voltmeter.
The circuit has two parts, one in the form of electrical wire which allows the flow of electrons and the other in the form of two solutions joined by salt bridge. In solution part of the circuit, the electric current is carried by movement of ions.

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Question 74 Marks

Identify whether the following reaction is redox or NOT. State oxidant and reductant therein.$3 H _3 AsO _{3( aq )}+ BrO _{5( qq )}^{-} \longrightarrow Br _{\text {(aq) }}^{-}+3 H _3 AsO _{4( aq )}$

Answer

i. Write oxidation number of all the atoms of reactants and products.

ii. Identify the species that undergoes change in oxidation number.

iii. The oxidation number of As increases from +3 to +5 and that of Br decreases from +5 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
iv. The oxidation number of As increases by loss of electrons and therefore, As is a reducing agent and itself is oxidised. On the other hand, the oxidation number of Br decreases and therefore, Br is an oxidising agent and itself is reduced by gain of electrons.
Result:
a. The given reaction is a redox reaction.
b. Oxidant/oxidising agent: $BrO _3^{-}$
c. Reductant/reducing agent: $H 3 AsO 3$

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Question 84 Marks

Justify the following reaction as redox reaction in terms of electron transfer. $Mg + F _2 \rightarrow MgF _2$

Answer

Redox reaction can be described as electron transfer as shown below:
$2Na_{(s)} + H_{2(g)} \rightarrow 2Na^+ + 2H^–$
$ii.$ Charge development suggests that each sodium atom loses one electron to form $Na^+$ and each hydrogen atom gains one electron to form $H^–.$ This can be represented as follows:

$iii. $When $Na$ is oxidised to $NaH,$ the neutral $Na$ atom loses one electron to form $Na^+$ in $NaH$ while the elemental hydrogen gains one electron and forms $H^–$ in $NaH.$
$iv.$ 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.

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Question 94 Marks

Explain redox reaction in terms of electron transfer.

Answer

$i.$ Redox reaction can be described in terms of electron transfer as shown below:
$2Mg_{(s)} + O_{2(g)} \rightarrow Mg^{2+} + 2O^{2-}$
$ii.$ Charge development suggests that each magnesium atom loses two electrons to form $Mg^{2+}$ and each oxygen atom gains two electrons to form $O^{2-}.$ This can be represented as follows:

$iii.$ When $Mg$ is oxidised to $MgO,$ the neutral $Mg$ atom loses electrons to form $Mg^{2+} \ in \ MgO$ while the elemental oxygen gains electrons and forms $O^{2-} \ in \ MgO.$
$iv.$ 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.

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Question 104 Marks

What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–

Answer

a. $Mg / Mg ^{2+}$
Here, $Mg$ loses two electrons to form $Mg ^{2+}$ ion. $Mg _{( s )} \longrightarrow Mg _{( aq )}^{2+}+2 e ^{-}$ Hence, $Mg / Mg ^{2+}$ is an oxidized state.

b. $Cu / Cu ^{2+}$
Here, Cu loses two electrons to form $Cu ^{2+}$ ion. $Cu _{( s )} \longrightarrow Cu _{( aq )}^{2+}+2 e ^{-}$
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. $O _{2( g )}+4 e ^{-} \longrightarrow 2 O _{( aq )}^{2-}$
Hence, $O _2 / O ^{2-}$ is in a reduced state.

d. $Cl _2 / Cl ^{-}$
Here, each $Cl$ gains one electron to form $Cl ^{-}$ion. $Cl _{2( g )}+2 e ^{-} \longrightarrow 2 Cl _{( aq )}^{-}$
Hence, $Cl _2 / Cl ^{-}$is in a reduced state.

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Question 114 Marks

How will you write Stock notations for the following compounds?
i. AuCl₃
ii. SnCl₄
iii. SnCl₂
iv. MnO₂

Answer

i. $AuCl _3$ : The charge on each element is $Au ^{3+} Cl _3^{1-}$. Hence, the stock notation is $Au ( III ) Cl _3$.
ii. $SnCl _4$ : The charge on each element is $Sn ^{4+} Cl _4^{1-}$. Hence, the stock notation is $Sn ( IV ) Cl _4$.
iii. $SnCl _2$ : The charge on each element is $Sn ^{2+} Cl _2^{1-}$. Hence, the stock notation is $Sn ( II ) Cl _2$
iv. $MnO _2$ : The charge on each element is $Mn ^{4+} O _2^{2-}$. Hence, the stock notation is $Mn ( IV ) O _2$.

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