Question

In which of the following reactions conc. $H_{2}S{O}_4$ is used as an oxidizing reagent?

• A. $2HI+H_{2}S{O}_4$
  
  $\downarrow$
  
  $I_2+S{O}_2+2H_{2}O$
  
• B. $Ca{F}_2+H_{2}S{O}_4$
  
  $\downarrow$
  
  $CaS{O}_4+2HF$
• C. $NaCl+H_{2}S{O}_4$
  
  $\downarrow$
  
  $NaHS{O}_4+HCl$
• D. $Cu+H_{2}S{O}_4$
  
  $\downarrow$
  
  $CuS{O}_4+S{O}_2+2H_{2}O$
  

Answer 1

Answer: (A), (D)

$Cu+H_{2}S{O}_4$


$\downarrow$

$CuS{O}_4+S{O}_2+2H_{2}O$

Analyzing the given options

Calculate the Oxidation number of S in each reaction.

Option (A):

$Ca{F}_2+H_{2}S{O}_4\to CaS{O}_4+HF$
The oxidation state of S is $H_{2}S{O}_4$ and $CaS{O}_4$ is the same i.e., +6.

It is not a redox reaction.

Option (B):

$2HI+H_{2}O\to I_2+S{O}_2+2H_{2}O$

Here the oxidation number of S in $H_{2}S{O}_4$ is +6 while in SO_2 it is +4.
Since the Oxidation number of S decreases in the given reaction,it is acting as an oxidizing agent.

$Cu+H_{2}S{O}_4\to CuS{O}_4+S{O}_2+2H_{2}O$

Here the oxidation number of $Sin{H}_{2}S{O}_4$ is +6, the oxidation number of $SinS{O}_2$ is +4.
Here, $H_{2}S{O}_4$ is acting as an oxidising agent.

Option (D):
$NaCl+H_{2}S{O}_4\to NaHS{O}_4+HCl$
Here, the Oxidation number of S in both the compound $H_{2}S{O}_4\text{and}NaHS{O}_4$ is the same i.e + 6

Thus, (B) and (C) are the correct options.