Redox Reactions

The strength of an aqueous $N a O H$ solution is most accurately determined by titrating : (Note : consider that an appropriate indicator is used)

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Raj Pandey

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Aq. $N a O H$ in a burette and aqueous oxalic acid in a conical flask.

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The redox reaction among the following is

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Contributor-Level 10

$N_{2} + O_{2} ? 2 N O$

Oxidation state changes.

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Oxidation number of potassium in $K_{2} O, K_{2} O_{2}$ $_{}_{}_{}$ and ${K O}_{2}$ $_{}$ respectively is

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Contributor-Level 10

$= \frac{10^{14}}{10^{3}} = 10^{11}$

$\underset{2 x - 2 = 0}{K_{2} O ? x = + 1}$

$\underset{2 x - 2 = 0}{K_{2} O_{2}} ? x = + 1$

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Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction:
CH? + I? ? CH?I + HI (Reversible)

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Contributor-Level 10

CH? + I? CH? I + HI
It is reversible due to reducing nature of HI but it can be irreversible by addition of oxidizing agents like, HNO? , HIO? , HgO, etc. which will oxidize HI into I?

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

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Contributor-Level 10

CrO? ²? (Cr? ) → Cr? ³, Total oxidation number change = 3
MnO? (Mn? ) → Mn? ², Total oxidation number change = 5
Cr? O? ²? (Cr? ) → 2Cr? ³, Total oxidation number change = 6
C? O? ²? (C? ³) → 2CO? (C? ), Total oxidation number change = 2

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In basic medium $C r O_{4}^{2 -}$ oxidises $S_{2} O_{3}^{2 -}$ to form $S O_{4}^{2 -}$ and itself changes into Cr(OH)₄^-. The volume of 0.154M $C r O_{4}^{2 -}$ required to react with 40mL of 0.25 M $S_{2} O_{3}^{2 -} i s$ -------------- mL.

(Rounded – off to the nearest integer)

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Contributor-Level 10

$\begin{array}{l} (+ 6) (+ 2) (+ 3) (+ 6) \\ C r {O_{4}}^{- 2} + \underset{}{\underset{?}{S_{2} {O_{3}}^{- 2} \to C r {(O H)}_{4}^{-} + S {O_{4}}^{- 2}}} \end{array}$

O.N.C = 4

n factor of S₂O₃^-- = 8

n factor of CrO₄^-- = 3

$m . e C r O_{4}^{- 2} = m . e . S_{2} O_{3}^{- 2}$

$V * (0.154 * 3) = 40 * (0.25 * 8) ∴ v = \frac{80}{0.462}$

= 173.16 ml $\approx$ 173 ml

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2 months ago

Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.

6 OH^- + Cl^- $C l O_{3}^{-}$ -> + 3H₂O + 6e^-

A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is__________. (Nearest Integer)

(Molar mass of KClO₃ = 122.6g mol^-1, F = 96500C)

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Contributor-Level 10

$6 O H^{-} + C l^{-} \to C l O_{3}^{-} + 3 H_{2} O + 6 e^{-}$

Mole of $C l O_{3}^{-} o r K C l O_{3} = \frac{10}{122.6} = 0.8456 m o l$

1 mole of $C l O_{3}^{-}$ or KClO₃ produced by 6 faraday (F).

$∴ 0.08456$ mole of $C l O_{3}^{-}$ or KClO₃ produced by (6 * 0.08156) F

$\frac{l * t}{F} = 6 * 0.8156$

$l = \frac{6 * 0.08156 * 965000}{10 * 60 * 60} A = 1.311 A \approx 1 A$

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2 months ago

Redox Reactions Class 12th

When 10 mL of an aqueous solution of KMnO₄ was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO₄ is grams per litre is________* 10^-2. (Nearest integer)

[Atomic mass of K = 39, Mn = 55, O = 16]

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Contributor-Level 10

Let molarity of KMnO₄ = x

$K M n O_{4} + F e S O_{4} \to F e_{2} {(S O_{4})}_{3} + M n^{2 +}$

n = 5 n = 1 Ferric sulphate

equivalent of KMnO₄ = equivalent of FeSO₄

5 * x * 10 = 1 * 0.1 * 10

x = 0.02 M

Strength = (0.02 * 158) = 3.16g/L

= 316 * 10^-2 g/L

Ans. = 316

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2 months ago

Redox Reactions Class 12th

The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is

$2 K_{2} C r_{2} O_{7} + 8 H_{2} S O_{4} + 3 C_{2} H_{6} O \overset{}{\to} 2 C r_{2} {(S O_{4})}_{3} + 3 C_{2} H_{4} O_{2} + 2 K_{2} S O_{4} + 11 H_{2} O$

If the rate of appearance of Cr₂(SO₄)₃ is 2.67 mol min^-1 at a particular time, the rate of disappearance of C₂H₆O at the same time is________mol min^-1. (Nearest integer)

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$2 K_{2} C r_{2} O_{7} + 8 H_{2} S O_{4} + 3 C_{2} H_{6} O \to 2 C r_{2} {(S O_{4})}_{3} + 3 C_{2} H_{4} O_{2} + 2 K_{2} S O_{4} + 11 H_{2} O$

Given that, rate of appearance of Cr₂(SO₄)₃ = $+ \frac{d [C r_{2} {(S O_{4})}_{3}]}{d t} = 2.67 m o l e / m i n$

LHS RHS

$(\frac{R a t e o f d i s a p p e a r a n c e o f C_{2} H_{6} O}{3}) = (\frac{R a t e o f a p p e a r a n c e o f C r_{2} {(S O_{4})}_{3}}{2})$

Rate of disappearance of C₂H₆O = $\frac{2.67 * 3}{2} m o l / m i n$

Rate of disappearance of C₂H₆O = 4.005 mol /min

Ans. = 4 (nearest integer)

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2 months ago

The redox reaction among the following is

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