1

### JEE Advanced 2020 Paper 2 Offline

MCQ (More than One Correct Answer)
Choose the correct statement(s) among the following :
A
SnCl2 . 2H2O is a reducing agent.
B
SnO2 reacts with KOH to form K2[Sn(OH)6].
C
A solution of PbCl2 in HCl contains Pb2+ and Cl$$-$$ ions.
D
The reaction of Pb3O4 with hot dilute nitric acid to give PbO2 is a redox reaction.

## Explanation

Sn2+ of stannous chloride dihydrate (SnCl2 . 2H2O) tends to convert into Sn4+.

Hence, statement (a) is correct.

(b) SnO2 reacts with KOH and gives K2SnO3 . 3H2O or K2[Sn(OH)6] because it is amphoteric in nature.

$$Sn{O_2} + KOH\buildrel {} \over \longrightarrow {K_2}Sn{O_3} + {H_2}O$$

or $${K_2}[Sn{(OH)_6}]$$

Hence, statement (b) is correct.

(c) In conc. HCl, PbCl2 exists as chloroplumbous acid, H2[PbCl4]

$$\mathop {PbC{l_2}}\limits^{II} + 2HCl\buildrel {} \over \longrightarrow \mathop {{H_2}[PbC{l_4}]}\limits^{II}$$

Hence, statement (c) is incorrect.

(d) Pb3O4 is a mixture of $$(2\mathop {PbO}\limits^{ + 2} + {\mathop {PbO}\limits^{ + 4} _2})$$

$$P{b_3}{O_4} + 4HN{O_3}\buildrel {} \over \longrightarrow \mathop {2Pb{{(N{O_3})}_2}}\limits^{ + 2} + \mathop {Pb{O_2}}\limits^{ + 4} + 2{H_2}O$$

It is not a redox reaction. Thus, the statement (d) is incorrect.
2

### JEE Advanced 2015 Paper 1 Offline

MCQ (More than One Correct Answer)

Fe3+ is reduced to Fe2+ by using

A
H2O2 in presence of NaOH.
B
Na2O2 in water.
C
H2O2 in presence of H2SO4.
D
Na2O2 in presence of H2SO4.

## Explanation

Fe3+ is reduced to Fe2+ by H2O2/NaOH and Na2O2/H2O.

$$2F{e^{3 + }} + {H_2}{O_2} + O{H^ - } \to 2F{e^{2 + }} + 2{H_2}O + {O_2}$$

$$N{a_2}{O_2} + {H_2}O \to {H_2}{O_2} + NaOH$$

3

### JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
For the reaction,

$${I^ - } + ClO_3^ - + {H_2}S{O_4} \to C{l^ - } + HSO_4^ - + {I_2}$$

the correct statement(s) in the balanced equation is/are
A
stoichiometric coefficient of HSO$$_4^ -$$ is 6
B
iodide is oxidised
C
sulphur is reduced
D
H2O is one of the products

## Explanation

Oxidation half-reaction,

$$2{I^ - } \to {I_2} + 2{e^ - }$$ .... (i)

Here, I$$-$$ is converted into I2. Oxidation number of I is increasing from $$-$$1 to 0 hence, this is a type of oxidation reaction.

Reduction half-reaction

$$6{H^ + } + ClO_3^ - + 6{e^ - } \to C{l^ - } + 3{H_2}O$$ ..... (ii)

Here, H2O releases as a product. Hence, option (d) is correct.

Multiplying equation (i) by 3 and adding in equation (ii)

Stoichiometric coefficient of $$HSO_4^ -$$ is 6.

Hence, option (a), (b) and (d) are correct.

#### Questions Asked from Redox Reactions

On those following papers in MCQ (Multiple Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
JEE Advanced 2020 Paper 2 Offline (1)
JEE Advanced 2015 Paper 1 Offline (1)
JEE Advanced 2014 Paper 1 Offline (1)

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