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Test: Redox reactions & Equivalent Weight (11 August) - JEE MCQ


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10 Questions MCQ Test - Test: Redox reactions & Equivalent Weight (11 August)

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Test: Redox reactions & Equivalent Weight (11 August) - Question 1

The decomposition of hydrogen peroxide to form water and oxygen is an example of

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 1

A disproportionation reaction is a particular type of redox reaction in which a single compound is transformed into two different compounds, one being reduced and the other being oxidized.

The decomposition of hydrogen peroxide to form water and oxygen is a classic example of a disproportionation reaction.

Test: Redox reactions & Equivalent Weight (11 August) - Question 2

Oxidation number denotes the oxidation state of an element in a compound ascertained on the basis that electron in a covalent bond belongs

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 2

The oxidation number is a concept in chemistry that helps us to keep track of electrons in an atom. It is based on the assumption that the electrons in a covalent bond belong entirely to the more electronegative element. This is a simplifying assumption that allows us to track electron movement and make predictions about chemical reactions.

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Test: Redox reactions & Equivalent Weight (11 August) - Question 3

Identify the correct statements with reference to the given reaction 

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 3

P4 is undergoing oxidation as well as reduction. As oxidation number of P4 is 0 in reactant and it increases to +3 in H2PO2and decreases to -3 in PH3

Test: Redox reactions & Equivalent Weight (11 August) - Question 4

The highest value of oxidation number changes from 1 to 7

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 4

Explanation of Highest Value of Oxidation Number Changes from 1 to 7

The highest value of oxidation number changes from 1 to 7 across the third period in the periodic table. This is due to the following reasons:

  • The Atoms of Transition Elements:Transition metals are those elements located in the d-block of the periodic table. They have varying oxidation states, but they do not usually reach an oxidation state of 7. Their oxidation states primarily range between +2 and +3, although some can reach states of +4 or +5.
  • The First Three Groups:The first three groups of the periodic table include alkali metals, alkaline earth metals, and boron group elements. These groups generally have oxidation states of +1, +2, and +3 respectively. They do not reach an oxidation state of 7.
  • In Alkaline Earth Metals:Alkaline earth metals are the elements in the second group of the periodic table. These elements generally have an oxidation state of +2 due to the presence of two valence electrons which are readily lost in chemical reactions.
  • Across the Third Period in the Periodic Table:The elements in the third period of the periodic table show a wider range of oxidation states, which can vary from +1 to +7. This is due to the presence of both s and p orbitals in their valence shell, which allows the elements to lose or gain more electrons in chemical reactions. Therefore, across the third period in the periodic table, the highest value of oxidation number changes from 1 to 7.
Test: Redox reactions & Equivalent Weight (11 August) - Question 5

The more positive the value of E0, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. 

E0values : Fe3 + / Fe2+ = +0.77; I2(s)/l- = +0.54; cu2+/ Cu = +0.34; Ag+ / Ag = +0.80V

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 5

The standard electrode potential, E0, is a measure of the tendency of a chemical species to be reduced. The higher the E0 value, the greater the tendency for reduction.

In a redox reaction, the chemical species that gets reduced acts as the oxidising agent. Therefore, the species with the highest E0 value will be the strongest oxidising agent.

Given the E0 values:

  • Fe3+ / Fe2+ = +0.77V
  • I2(s) / I- = +0.54V
  • Cu2+ / Cu = +0.34V
  • Ag+ / Ag = +0.80V


From the given values, it is clear that Ag+ / Ag has the highest E0 value (+0.80V). Therefore, Ag+ is the strongest oxidising agent among the given options.

Conclusion:

  • Ag+ is the strongest oxidising agent because it has the highest standard electrode potential (E0) value among the given species.
  • Remember, the higher the E0 value, the greater the tendency of the species to be reduced, and hence, stronger is its oxidising power.
Test: Redox reactions & Equivalent Weight (11 August) - Question 6

Consider conversion of MnCI2 into

      
Increasing order of equivalent weights of MnCI2 in these conversions is

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Test: Redox reactions & Equivalent Weight (11 August) - Question 7

The number of moles of KMnO4 that will be needed to react with one mole of ferrous oxalate Fe(C2O4) in acidic solution is

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 7

Test: Redox reactions & Equivalent Weight (11 August) - Question 8

Photosynthesis of carbohydrates in plants takes place as

6CO2 +12H2 C6H12O6 + 6O2 + 6H2O

Equivalent weight of CO2 and C6H12O6 respectively are 

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 8


Change in oxidation number = 24 units

Test: Redox reactions & Equivalent Weight (11 August) - Question 9

Equivalent w eight of KMnO4 and its reduced species in different mediums are given


Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 9



 

*Multiple options can be correct
Test: Redox reactions & Equivalent Weight (11 August) - Question 10

In which of the following reactions, equivalent mass of the underlined is equal to molar mass?

Detailed Solution for Test: Redox reactions & Equivalent Weight (11 August) - Question 10





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