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Chemistry 2 Marks Questions

PART - 2 CH - 7 Redox Reactions · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 13 questions.

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Question 12 Marks
Name the compound in which :
(i) Oxidation number of oxygen is +2 .
(ii) Oxidation number of oxygen is +1 .
(iii) Oxidation number of chlorine is +1 .
Answer
(i) $OF _2$ (ii) $H _2 O _2$ (iii) ClF
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Question 22 Marks
How standard electrode potential can be calculated and state its importance.
Answer
We can't calculate the electrode potential of a single electrode. Hence it can be calculated by combining it with standard electrode.
Standard electrode potential or Standard reduction potential $=$ Electrode potential of cathode - Electrode potential of anode
$E _{\text {cell }}^0= \underset{\text{cathode}}{E ^0}- E _{\text {anode }}^0$
The electrode potential of hydrogen electron is considered at zero in standard conditions. Hence, when it is attached with unknown electrode, the electrode potential can be calculated.
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Question 32 Marks
(a) Name the halide ion whose oxidation can't be done by chemical method. How it can be oxidised?
(b) Explain the bleaching action of chlorine.
Answer
(a) The oxidation of $F ^{-}$(fluoride ion) can't be done by chemical method because electronegativity of fluorine is maximum the oxidation of fluoride ion is done by electrical decomposition.
(b) In presence of moisture, Cl behave as bleach. From this $ClO ^{-}$and $Cl ^{-}$ion get form. Hypochlorite ion oxidise the colour and make it colourless.
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Question 42 Marks
Balance the following reaction by ion electron method (half-reaction method).
$MnO _4^{-}( aq )+ H ^{+}( aq )+ Fe ^{2+}( aq ){+ H _2 O (l)}{\longrightarrow} Mn ^{2+}( aq )+Fe ^{3+}( aq )$
Answer
This is a redox reaction which can be written as two half reactions in following ways :
Oxidation: $Fe ^{2+} \longrightarrow Fe ^{3+}$
Reduction : $MnO _4^{-} \longrightarrow Mn ^{2+}$
On balancing half reactions : $Fe^{2+} \longrightarrow Fe^{3+}+e^{-} \text {(Charge balance) } \quad \ldots(1)$
$MnO _4^{-} \longrightarrow Mn ^{2+}+{4 H _2 O }\text { (Oxygen balancing) }$
$MnO_4^{-}+8 H^{+} \longrightarrow Mn^{2+}+{4 H_2 O} \text{(Balancing hydrogen)}$
$MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+4 H_2 O\text{(Charge balancing)...(2)}$
On multiplying 5 to equation (1) and adding to equation (2),
Image
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Question 52 Marks
What do you meant by Redox titration? Explain.
Answer
Those titrations in which one substance get oxidised while other get reduced is called redox titration. In these titrations Potassium permanganate,
Potassium dichromate $\left( KMnO _4, K_2 Cr _2 O _7\right)$ etc. are used as oxidizing agents while ferrous sulphate, ferrous ammonium sulphate and oxalic acid etc. are used as reducing agent. The $KMnO _4$ is filled in burette and it work as self indicator. Hence, indicator is not required and at last point pink colour will appear. In other conditions diphenyl amine, N -phenyl anthranilic acid etc. work as indicators.
In titration of $K _2 Cr _2 O _7$, diphenylamine indicator get oxidised and at last point give blue colour. By these titrations we calculate the strenghs of oxidants and reductants.
The titration between $CuSO _4$ (copper sulphateoxidizing agent) and $Na _2 S_2 O _3$ (sodium thiosulphatereducing agent) is a prime example of redox titration. In this titration KI is used due to which the $I ^{-}$obtained get oxidized by $Cu ^{+2}$ ions and gets converted into $I _2$.
$2 Cu ^{2+}( aq )+4 I ^{-}( aq ) \longrightarrow Cu _2 I _2( aq )+ I _2( aq )$
Here the starch indicator is used which gives blue colour with $I _2$. Then $I _2$ reacts with $S _2 O _3{ }^{-2}$ ion and at end point the solution becomes colourless.
$I _2( aq )+2 S_2 O _3{ }^{2-}( aq ) \longrightarrow 2 I ^{-}( aq )+ S _4 O _6{ }^{2-}( aq )$
$I _2$ is insoluble in water. Hence in solution of KI, it remain as solute in form of $KI _3$.
In this titration, iodine has been produced in the reaction instead of being taken directly, hence this type of titration are called iodometric titrations and those titrations in which iodine solution is taken directly and its action is done with sodium thiosulphate, then it is called iodimetric titration.
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Question 72 Marks
Calculate the oxidation states of Carbon, Bromine and Manganese in $C _3 O _2$ (Carbon suboxide), $Br _3 O _8$ (Tribromo octaoxide) and $Mn _3 O _4$ respectively.
Answer
(i) $C _3 O _2$ (Carbon suboxide)
In $[ O =\stackrel{+2}{ C }=\stackrel{0}{ C }=\stackrel{+2}{ C }= O ]$, according to the structure the oxidation states of carbon atoms is $+2,0$ and +2 respectively. Hence, its average oxidation state is $\frac{4}{3}$.
(ii) The structure of $Br _3 O _8$ is as follows :
Image
$\quad\quad\quad$According to this the oxidation states of different bromine atoms is $+6,+4$ and +6 respectively. Hence its average oxidation number is $\frac{+16}{3}$.
(iii) $Mn _3 O _4$ is mixed oxide which is formed by the combination of two MnO and one $MnO _2$ in which oxidation number of Mn is +2 and +4 respectively. Hence, the average oxidation state of Mn is $\frac{8}{3}$.
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Question 82 Marks
Calculate the oxidation number :
(i) Br in HOBr
(ii) Cl in $KClO _3$
(iii) Cl in ICl
(iv) C in $\overline{ C } N$
(v) $P$ in $NaHPO _2$
(vi) Fe in $Fe _{0.94} O$
Answer
(i)
$\begin{array}{l} HOBr \\ +1-2+x=0 \\ x=+1\end{array}$
(ii)
$\begin{array}{l} KClO _3 \\ +1+x-2(3)=0 \\ x=+5\end{array}$
(iii)
$\begin{array}{l} ICl \\ +1+x=0 \\ x=-1\end{array}$
(iv)
$\begin{array}{l}\overline{ C } N \\ x-3=-1 \\ x=+2\end{array}$
(v)
$\begin{array}{l} NaHPO _2 \\ +1+1+x-2(2)=0 \\ x=+2\end{array}$
(vi)
$\begin{array}{l} Fe _{0.94} O \\ 0.94 x-2=0 \\ 0.94 x=2 \\ x=\frac{2}{0.94}\end{array}$
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Question 92 Marks
Identify the oxidising and reducing agent in following chemical reactions :
(i) $2 M g + S O _2 \longrightarrow 2 M g O + S$
(ii) $2 Cu ^{2+}+4 I ^{-} \longrightarrow 2 CuI + I _2$
(iii) $SO _2+2 H _2 S \longrightarrow 2 H _2 O +3 S$
(iv) $S n ^{2+}+ 2 H g ^{2+} \longrightarrow H g _2{ }^{2+}+ Sn ^{4+}$
Answer
(i) $SO _2$ is an oxidant while Mg is reductant.
(ii) $Cu ^{2+}$ is an oxidant while $I ^{-}$is reductant.
(iii) $SO _2$ is an oxidant while $H _2 S$ is reductant.
(iv) $Hg ^{2+}$ is an oxidant while $Sn ^{+2}$ is reductant.
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Question 102 Marks
Explain stock notation with example.
Answer
According to scientist Alfred Stock, the oxidation number of metals in compounds is written in bracket in Roman numerals, this is known as stock notation.
Example : Aurous chloride can be written as $Au ( I ) Cl$ and auric chloride can be written as Au (III) $Cl _3$. Similiarly stannous chloride can be written as Sn (II) $Cl _2$ and stannic chloride can be written as $Sn ( IV ) Cl _4$.
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Question 112 Marks
Explain the following with reason :
(i) On the iron nails dipped in $CuSO _4$ solution, Cu get deposited.
(ii) Na reacts with water at normal temperature while Mg react with hot water.
Answer
(i) Fe is more reactive than Cu . As more reactive metal, displaces less reactive metal from its salt solution. Hence, following reaction takes place :
$Fe(s)+CuSO_4(aq) \longrightarrow FeSO_4(aq)+Cu(s)$
The released copper get deposited on iron nails.
(ii) Na is more reactive than Mg . Hence, Na react with water at normal temperature while Mg react with water at hot temperature.
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Question 122 Marks
In following reactions, which substance is getting oxidised and which is getting reduced?
(i) $Sn ^{2+}+2 Fe ^{3+} \longrightarrow Sn ^{4+}+ Fe ^{2+}$
(ii) $Zn + Fe ^{2+} \longrightarrow Z n ^{2+}+ Fe$
(iii) $2 Na + C l _2 \longrightarrow 2 NaCl$
(iv) $M g +2 H ^{+} \longrightarrow M g ^{2+}+ H _2$
Answer
(i) $Sn ^{+2}$ is getting oxidised while $Fe ^{+3}$ is getting reduced.
(ii) Zn is getting oxidised and $Fe ^{+2}$ is getting reduced.
(iii) Na is getting oxidised and $Cl _2$ is getting reduced.
(iv) Mg is getting oxidised and $H ^{+}$is getting reduced.
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Question 132 Marks
Give different definations of Reduction reactions.
Answer
Reduction reaction is a reaction in which :
(i) Any substance get added to hydrogen.
(ii) Electropositive element combine with it.
(iii) Removal of oxygen.
(iv) Addition of electrons in any species.
(v) Decrease in oxidation number.
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