Q: H 2 S ( 5 moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is x , and the number of moles of electrons in

KMnO 4 ⟶ H + Mn 2 + S - 2 ⟶ S 0 The number of electrons involved per molecule is called n-factor. n-factor of KMnO 4 = 5 n-factor of S - 2 H 2 S = 2 Moles × n - factor = number of equivalents n KMnO 4 × 5 = ( 5 × 2 ) H 2 S ∴ n KMnO 4 = 2 ∴ 2 KMnO 4 + 3 H 2 SO 4 + 5 H 2 S → K 2 SO 4 + 2 MnSO 4 + 5 S + 8 H 2 O Number of moles of water produced = 8 Number of moles of el

Q: A sample (5.6 ~g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 ~mL of solution. Titration of 25.0 ~mL of this solution requires 12.5 ~mL of 0.03 M KMnO _ 4 solution to

Moles of Fe 2 + present in 250 ml solution = x × 10 - 2 Moles of Fe 2 + present in 25 ml solution = x × 10 - 3 mole At equivalence point eq. of Fe 2 + = eq. of KMnO 4 Moles Fe 2 + × vf Fe 2 + = Moles KMnO 4 × vf KMnO 4 x × 10 − 3 × 1 = 0 .03 × 12 .5 × 10 − 3 × 5 x = 1 . 875 mole Moles of Fe 2 + present in 250 &

Q: A sample (5.6 ~g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 ~mL of solution. Titration of 25.0 ~mL of this solution requires 12.5 ~mL of 0.03 M KMnO _ 4 solution to

Moles of Fe 2 + present in 250 ml solution = x × 10 - 2 Moles of Fe 2 + present in 25 ml solution = x × 10 - 3 mole At equivalence point eq. of Fe 2 + = eq. of KMnO 4 Moles Fe 2 + × vf Fe 2 + = Moles KMnO 4 × vf KMnO 4 x × 10 − 3 × 1 = 0 .03 × 12 .5 × 10 − 3 × 5 x = 1 . 875 mole Moles of Fe 2 + present in 250 &

Q: In the chemical reaction between stoichiometric quantities of KMnO 4 KI in weakly basic solution, what is the number of moles of I 2 released for 4 KMnO 4 consumed?

2 MnO 4 4 mole + 6 I − → 3 I 2 6 mole + 2 MnO 2

Question 1

One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is

Accepted Answer

I ^ - +2 MnO _4^ - + H _2 O Neutral solution 2 MnO _2+ IO _3^ - +2 OH ^ -

Question 2

In a metal deficient oxide sample, M _ x Y _2 O _4 ( M and Y are metals), M is present in both +2 and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M ^ 2+ ions present in M is

Accepted Answer

Average oxidation state of M = 1 3 2+ 2 3 3=+ 8 3 aligned & For M _ X M _ 2 Y _4 \ & 8 3 x +3 2+4(-2)=0 \ & 8 3 x =2 \ & x = 3 4 =0.75 aligned

Question 3

At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid ( HNO _2 ) gives the species

Accepted Answer

3 HNO _2( aq ) H _3 O ^ + + NO _3^ - +2 NO

Question 4

H 2 &#160; S ( 5 moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is x , and the number of moles of electrons in

Accepted Answer

KMnO 4 &#10230; H + Mn 2 + S - 2 &#10230; &#160; S 0 The number of electrons involved per molecule is called n-factor. n-factor of KMnO 4 = 5 n-factor of S - 2 H 2 &#160; S = 2 Moles &#215; n - factor &#160; = &#160; number &#160; of &#160; equivalents   n KMnO 4 &#215; 5 = ( 5 &#215; 2 ) H 2 &#160; S &#8756; n KMnO 4 = 2 &#8756; 2 KMnO 4 + 3 H 2 SO 4 + 5 H 2 &#160; S &#8594; K 2 SO 4 + 2 MnSO 4 + 5 &#160; S + 8 H 2 O Number of moles of water produced = &#160; 8 &#160; Number of moles of el

Question 5

Consider the following molecules : Br 3 O 8 , &#160; F 2 O , H 2 &#160; S 4 O 6 , H 2 &#160; S 5 O 6 , and C 3 O 2 . Count the number of atoms existing in their zero oxidation state in each molec

Accepted Answer

The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. When same atoms are connected, there is no electronegativity difference, hence, no exchange of electrons. Hence, they have zero oxidation states. The structures of the given molecules are shown below with their oxidation states. Total atom with zero oxidation number state are 6 .

Question 6

A sample (5.6 ~g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 ~mL of solution. Titration of 25.0 ~mL of this solution requires 12.5 ~mL of 0.03 M KMnO _ 4 solution to

Accepted Answer

Moles of  Fe 2 +  present in  250 &#160; ml  solution  = x &#215; 10 - 2 Moles of  Fe 2 +  present in  25 &#160; ml  solution  = &#160; x &#215; 10 - 3 &#160; mole At equivalence point eq. of  Fe 2 +  = eq. of  KMnO 4 Moles Fe 2 + &#215; vf Fe 2 + = Moles KMnO 4 &#215; vf KMnO 4 x &#215; 10 &#8722; 3 &#215; 1 = 0 .03 &#215; 12 .5 &#215; 10 &#8722; 3 &#215; 5 x = 1 . 875 &#160; mole Moles of  Fe 2 +  present in  250 &

Question 7

A sample (5.6 ~g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 ~mL of solution. Titration of 25.0 ~mL of this solution requires 12.5 ~mL of 0.03 M KMnO _ 4 solution to

Accepted Answer

Moles of  Fe 2 +  present in  250 &#160; ml  solution  = x &#215; 10 - 2 Moles of  Fe 2 +  present in  25 &#160; ml  solution  = &#160; x &#215; 10 - 3 &#160; mole At equivalence point eq. of  Fe 2 +  = eq. of  KMnO 4 Moles Fe 2 + &#215; vf Fe 2 + = Moles KMnO 4 &#215; vf KMnO 4 x &#215; 10 &#8722; 3 &#215; 1 = 0 .03 &#215; 12 .5 &#215; 10 &#8722; 3 &#215; 5 x = 1 . 875 &#160; mole Moles of  Fe 2 +  present in  250 &

Question 8

In the chemical reaction between stoichiometric quantities of KMnO 4 KI in weakly basic solution, what is the number of moles of I 2 released for 4 KMnO 4 consumed?

Accepted Answer

2 MnO 4 4 &#8196; mole + 6 I &#8722; &#8594; 3 I 2 6&#8196;mole&#160; + 2 MnO 2

Question 9

An acidified solution of potassium chromate was layered with an equal volume of amyl alcohol. When it was shaken after the addition of 1 &#160; mL of 3 % &#160; H 2 O 2 , a blue alcohol layer was obta

Accepted Answer

K 2 CrO 4 → H + / H 2 O 2 CrO 5 ( Deep Blue solution ) No. of oxygen atom bonded with chromium with single bond is ( 4 ) .

Question 10

Choose the correct statement(s) among the following.

Accepted Answer

(A)  Sn 2 +  is a good reducing agent which gets oxidise into  Sn 4 + (B) Amphoteric nature. (C)  PbCl 2 + 2 Cl - &#10230; PbCl 4 2 - (D)  Pb 3 O 4 ( 2 PbO &#8196; + &#8196; PbO 2 ) + 4 HNO 3 ( Conc . ) &#8594; PbO 2 &#8595; + 2 Pb NO 3 2 + 6 H 2 O

Question 11

5 . 00 &#160; mL of 0 . 10 &#160; M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH  required for the app

Accepted Answer

True data for NaOH used volume = 9 ml Milli equivalent of H 2 C 2 O 4 = Milli equivalent of NaOH 5 &#215; 0 . 1 &#215; 2 = 9 &#215; &#215; M &#215; 1 x = 1 9 = 0 . 11 M

Redox Reactions — Previous Year Questions & Solutions