Question

Match the LIST-I with LIST-II (Redox Reactions).

Choose the correct answer from the options given below:

• A. A-IV, B-I, C-II, D-III
• B. A-II, B-III, C-I, D-IV
• C. A-II, B-III, C-IV, D-I
• D. A-III, B-IV, C-I, D-II

Hint:

In redox reactions, keep track of oxidation states to determine the type of reaction (combination, displacement, etc.).

Answer 1

The Correct Option is C

Let's analyze each reaction:

A. $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$:
This is a combustion reaction where methane reacts with oxygen to produce carbon dioxide and water.
The oxidation state of carbon changes from -4 to +4, and the oxidation state of oxygen changes from 0 to -2.
This is also a combination reaction.
So, A is matched with II.

B. $2NaH(s) \rightarrow 2Na(s) + H_2(g)$:
Sodium hydride decomposes into sodium and hydrogen.
The oxidation state of sodium changes from +1 to 0, and the oxidation state of hydrogen changes from -1 to 0.
This is a decomposition reaction.
So, B is matched with III.

C. $V_2O_5(s) + 5Ca(s) \rightarrow 2V(s) + 5CaO(s)$:
This is a displacement reaction, where calcium displaces vanadium from its oxide.
Calcium is oxidized from 0 to +2, and vanadium is reduced from +5 to 0.
So, C is matched with IV.

D. $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$:
Hydrogen peroxide decomposes into water and oxygen.
The oxidation state of oxygen in $H_2O_2$ is -1, in $H_2O$ it is -2, and in $O_2$ it is 0.
This is a disproportionation reaction, where oxygen in $H_2O_2$ is both oxidized and reduced.
So, D is matched with I.

Final Matching:
A - II
B - III
C - IV
D - I

Final Answer:
The final answer is $ A\text{-}II,\ B\text{-}III,\ C\text{-}IV,\ D\text{-}I $.