Redox Reactions Questions — JEE Chemistry

Q1

Consider the following reaction:

xMnO_4^- + yC_2O_4^{2-}

+ zH+

\to

xMn2+ + 2yCO2 +

{z \over 2}{H_2}O

The value's of x, y and z in the reaction are, respectively :

A 5, 2 and 16

B 2, 5 and 8

C 2, 5 and 16

D 5, 2 and 8

Correct Answer

Option C

Solution

After balancing the reaction we get

2MnO_4^- + 5C_2O_4^{2-}

+ 16H+ $\to$ 2Mn2+ + 10CO2 +

8{H_2}O

$\therefore$

x

= 2,

y

= 5 and z = 16

Q2

An example of a disproportionation reaction is :

A 2NaBr + Cl2

\to

2NaCl + Br2

B 2KMnO4

\to

2KMnO4 + MnO2 + O2 (3) (4)

C 2CuBr

\to

CuBr2 + Cu

D 2MnO4 + 10I– + 16H+

\to

2Mn2+ + 5I2 + 8H2O

Correct Answer

Option C

Solution

In disproportionation reaction one element undergoes both oxidation and reduction. Here disproportionation reaction is :

Q3

In alkaline medium,

\mathrm{MnO}_4^{-}

oxidises

\mathrm{I}^{-}

to

A

I_2

B

\mathrm{IO}_3^{-}

C

\mathrm{IO}^{-}

D

\mathrm{IO}_4^{-}

Correct Answer

Option B

Solution

2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O}+\mathrm{I}^{-} \xrightarrow{\text{ alkaline medium }} 2 \mathrm{MnO}_2+2 \mathrm{OH}^{-}+\mathrm{IO}_3^{-}

$

Q4

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

. Assertion A: Phenolphthalein is a

\mathrm{pH}

dependent indicator, remains colourless in acidic solution and gives pink colour in basic medium. Reason R: Phenolphthalein is a weak acid. It doesn't dissociate in basic medium. In the light of the above statements, choose the most appropriate answer from the options given below.

A Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

.

B Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

.

C A is true but R is false.

D A is false but R is true.

Correct Answer

Option C

Solution

Phenolphthalein is a pH dependent indicator.

It is a weak acid which is colourless in the acidic solution but gives pink colour in basic medium.

The pink colour is due to its conjugate form.

Therefore, assertion (A) is true but Reason (R) is false.

Q5

The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH− in aqueous solution is :

A 200 mL

B 400 mL

C 600 mL

D 800 mL

Correct Answer

Option B

Solution

According to law of equivalence, Equivalence of acid = Equivalence of base.

Equivalence of acid = Normality x volume = 0.1 $\times$ v As we know base produce OH $-$ ion, so moles of base is same as moles of OH $-$ ion = 0.04 Another formula of equivalence = n factor $\times$ number of moles

\therefore\,\,\,

Equivalance of base = n factor of OH $-$ $\times$ moles of OH $-$ = 1 $\times$ 0.04 As for any ion, the charge of that ion is the n factor of that ion.

Here OH $-$ has 1 negative charge so it's n factor = 1

\therefore\,\,\,

0.1 $\times$ v = 1 $\times$ 0.04 $\Rightarrow$

\,\,\,

v = 0.4 L = 0.4 $\times$ 1000 = 400 ml.

Q6

Oxidation number of potassium in K2O. K2O2 and KO2 respectively is :

A +1, +2 and + 4

B +1, +1 and + 1

C +2, +1 and

+ {1 \over 2}

D +1, +4, and +2

Correct Answer

Option B

Solution

Potasisum has an oxidation of +1 (only) in combined state.

Q7

Identify the process in which change in the oxidation state is five :

A

C{r_2}O_7^{2 - } \to 2C{r^{3 + }}

B

MnO_4^ - \to M{n^{2 + }}

C

CrO_4^{2 - } \to C{r^{3 + }}

D

{C_2}O_4^{2 - } \to 2C{O_2}

Correct Answer

Option B

Solution

MnO_4^ - + 5e \to M{n^{ + 2}}

Q8

The compound that cannot act both as oxidising and reducing agent is :

A H3PO4

B H2SO3

C H2O2

D HNO2

Correct Answer

Option A

Solution

In H3PO4, P is present in +5 oxidation state and it can act as reducing agent only.

Q9

The hardness of a water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 is (Molar mass of CaSO4 = 136 g mol–1)

A 90 ppm

B 100 ppm

C 50 ppm

D 10 ppm

Correct Answer

Option B

Solution

nCaSO4 $\times$ Van't hoff factor = nCaCO3 $\times$ Van't hoff factor $\Rightarrow$ 10-3 $\times$ 2 = nCaCO3 $\times$ 2 $\Rightarrow$ nCaCO3 = 10-3 mol in 1 L $\therefore$ wCaCO3 = 100 $\times$ 10-3 g CaCO3 in 1 L solution $\therefore$ hardness in terms of CaCO3 =

{{{w_{CaC{O_3}}}} \over {{w_{Total}}}} \times {10^6}

=

{{100 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}

= 100 ppm

Q10

The oxidation states of 'P' in H4P2O7, H4P2O5 and H4P2O6, respectively, are :

A 7, 5 and 6

B 5, 4 and 3

C 5, 3 and 4

D 6, 4 and 5

Correct Answer

Option C

Solution

Oxidation state of P in H4P2O7, H4P2O5 and H4P2O6 is 5, 3 & 4 respectively H4P2O7 2x + 4(+ 1) + 7 ( $-$ 2) = 0 x = + 5

{H_4}\underline {{P_2}} {O_5}

2x + 4(+ 1) + 5 ( $-$ 2) = 0 x = + 3

{H_4}\underline {{P_2}} {O_6}

2x + 4 (+ 1) + 6( $-$ 2) = 0 x = +4