Redox Reactions MCQs for NEET — Chemistry Questions with Answers

Let normality of HCl is N₁ and H₂SO₄ is N₂.

∴ M.e. of HCl + M.e. of H₂SO₄ = M.e of NaOH

25 × N₁ + 25 × N₂ = 10 × 1 ... (1)

N₁ + N₂ = 0.4 ... (1)

BY POAC

Moles of Cl = moles of AgCl

$\frac{20\times {\text{\hspace{0.17em}N}}_{\text{1}}}{1000}=\frac{0.1435}{143.05}={10}^{-3}$

N₁ = 0.05 N

N₂ = 0.35 N 

6e^- + 10H⁺ + V₂O₅ → 2V2⁺ + 5 H₂ O

Zn → Zn²⁺ + 2e^- 3 × 3

V₂O₅ + 3 Zn + 10 H⁺ → 3Zn²⁺ + 2V²⁺ + 5H₂O    …

Now H₂O + V²⁺ → VO²⁺ + 2H⁺ + 2e^-

2e^- + I₂ → 2r^-

V²⁺ + I₂ + H₂O → 2I^- + VO²⁺ + 2H⁺

so we have 1 moles of V₂O₅ will reduce 2 moles of iodine

so $\left(\frac{{\displaystyle 10}}{{\displaystyle 102+80}}\right)\text{×\hspace{0.17em}2}$ moles of will be reduced by given amount of V₂O₅ = 0.11moles of 1₂

M.e. of NH₃ = 10

m.e. of NH3 = 10

m.mole of (NH₄)₂SO₄= 5

wt. of (NH₄)₂ SO₄ = 5

wt. of (NH4)2 SO4 = $\frac{5}{1000}\times$ 132 = 0.66 gm

% of (NH₄)₂ S0₄ = $\frac{0.66}{0.7}\times$ 100 = 94.28 %

KOH + Na₂aO₃

a M.e.b.M.e.

a + $\frac{\text{b}}{\text{2}}$ = 15 × $\frac{1}{20}$

2a + b= 1.5 … (i) (in presence of phenolphthalein)

a + b = 25 × $\frac{1}{20}$= 1.25… (ii) (in presence of Methyl orange)

by solving (i) & (ii) a = 0.25 m.e

mass of KOH = $\frac{0.25}{1000}\times$ 56 = 0. 014 gm

                                 AgNO₃    +    NaCl      →      AgCl   +   NaNO₃  

Moles before mixing $\frac{5.77}{108+14+48}$       $\frac{4.77}{58.5}$                   0             0

                                   = 0.0339        = 0.0815     (Here AgNO₃ is limiting reactant, thus)

Moles after mixing            0            0.0815 – 0.0339        0.0339       0.0339

                                                         = 0.0476

∴ Moles of AgCl formed = 0.0339

∴ Mass of AgCl formed = Mol. mass × No. of moles = 143.5 × 0.0339 = 4.864 gm

Let the oxidation state of nitrogen in the given compounds be x.

(a)  x  +1

      N₂H₄

              2x + (+1)4 = 0

                       2x = -4

                .^..     x = -2

(b)  x +1

      NH₃              x + (+1)3 =0

                          .^.. x = -3

(c)  x +1

      N₃H               x3 + (+1)0 = 0

                             3x = -1

                        .^.. x = -1/3

(d)  x+1-2+1

      NH₂OH              x + (+1)2+(-2)+(+1) =0

                                 x+2-2+1=0

                         .^.. x =-1

Key Idea Oxidation state of H is +1 and that of O is -2.

Let the oxidation state of P in the given compounds is x.

In H₄P₂O₅ .

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

                 4+ 2x-10 = 0

                 2x = 6

                 x = + 3

In H₄P₂O₆,

                     (+1)x4+2x + (-2)x6 = 0

                     4 + 2x-12=0

                     2x = 8

                     x = + 4

In H₄P₂O₇,

                            (+1)x4+2 x (x + (-2)x7)=0

                             4+2x -14 = 0

                              2x=10

                               x=+ 5

Thus, the oxidation states of P in H₄P₂O₅, H₄P₂O₆, H₄P₂O₇ are + 3 + 4 and +5 respectively.

Key Idea (i) Sum of oxidation states of all atoms = charge of ion.

(ii) Oxidation number of oxygen = -2

 Let the oxidation state of P in ${\mathrm{PO}}_4^{3-}$ be x.

                        ${\mathrm{PO}}_4^{3-}$