
The substances which may be oxidized as well as reduced are:
A. $HCl$
B. $HClO$
C. $HCl{O_3}$
D. $HCl{O_4}$
Answer
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Hint:The oxidizing and reducing power depends on the oxidation state of the element. In order to identify the substance which may go under oxidation and reduction from the following option we have to highlight the oxidation state of chlorine in each substance.
Complete answer:
From the above option the substances which may be oxidized as well as reduced are those which have capability to lose an electron as well to gain an electron.
As we know the possible oxidation state of chlorine are: $ - 1,0, + 1, + 3, + 5, + 7$
Now, in option A that is $HCl$ ,
Let oxidation state of chlorine be $x$
So, $x + 1 = 0$ (as oxidation state of hydrogen is $1$ )
$ \Rightarrow x = - 1$
So, the oxidation state of chlorine in $HCl$ is $ - 1$ .
Chlorine cannot accept more electrons in such cases. So, it cannot be reduced.
Now, in option B that is $HClO$ ,
Let oxidation state of chlorine be $x$
So, $1 + x + ( - 2) = 0$ (as oxidation state of hydrogen is $1$ and oxidation state of oxygen is $ - 2$ )
$ \Rightarrow x = 1$
So, the oxidation state of chlorine in $HClO$ is $1$ .
Here, chlorine can accept more electrons as well as loses more electrons. That means it gets reduced as well as oxidized.
Now, in option C that is $HCl{O_3}$ ,
Let oxidation state of chlorine be $x$
So, $1 + x + ( - 2) \times 3 = 0$ (as oxidation state of hydrogen is $1$ and oxidation state of oxygen is $ - 2$ )
$ \Rightarrow x = 5$
So, the oxidation state of chlorine in $HCl{O_3}$ is $5$ .
Here, chlorine can accept more electrons as well as loses more electrons. That means it gets reduced as well as oxidized.
Now, in option D that is $HCl{O_4}$ ,
Let oxidation state of chlorine be $x$
So, $1 + x + ( - 2) \times 4 = 0$ (as oxidation state of hydrogen is $1$ and oxidation state of oxygen is $ - 2$ )
$ \Rightarrow x = 7$
So, the oxidation state of chlorine in $HCl{O_4}$ is $7$ .
Chlorine cannot lose more electrons in such cases. So, it cannot be oxidized.
So, $HClO$ and $HCl{O_4}$ are the two substances among them which may be reduced as well as oxidized.
Hence, option B and C are correct.
Note:
Redox reactions are such types of reactions where oxidation and reduction both take place simultaneously. In redox reactions oxidation states of atoms get changed. In this reaction one species undergoes oxidation and on the other hand other species undergoes reduction. The oxidation half is identified by using the oxidation state such as increase in oxidation state or loss of electron is oxidation and the reduction half is identified where oxidation state decreases and gain of electrons takes place.
Complete answer:
From the above option the substances which may be oxidized as well as reduced are those which have capability to lose an electron as well to gain an electron.
As we know the possible oxidation state of chlorine are: $ - 1,0, + 1, + 3, + 5, + 7$
Now, in option A that is $HCl$ ,
Let oxidation state of chlorine be $x$
So, $x + 1 = 0$ (as oxidation state of hydrogen is $1$ )
$ \Rightarrow x = - 1$
So, the oxidation state of chlorine in $HCl$ is $ - 1$ .
Chlorine cannot accept more electrons in such cases. So, it cannot be reduced.
Now, in option B that is $HClO$ ,
Let oxidation state of chlorine be $x$
So, $1 + x + ( - 2) = 0$ (as oxidation state of hydrogen is $1$ and oxidation state of oxygen is $ - 2$ )
$ \Rightarrow x = 1$
So, the oxidation state of chlorine in $HClO$ is $1$ .
Here, chlorine can accept more electrons as well as loses more electrons. That means it gets reduced as well as oxidized.
Now, in option C that is $HCl{O_3}$ ,
Let oxidation state of chlorine be $x$
So, $1 + x + ( - 2) \times 3 = 0$ (as oxidation state of hydrogen is $1$ and oxidation state of oxygen is $ - 2$ )
$ \Rightarrow x = 5$
So, the oxidation state of chlorine in $HCl{O_3}$ is $5$ .
Here, chlorine can accept more electrons as well as loses more electrons. That means it gets reduced as well as oxidized.
Now, in option D that is $HCl{O_4}$ ,
Let oxidation state of chlorine be $x$
So, $1 + x + ( - 2) \times 4 = 0$ (as oxidation state of hydrogen is $1$ and oxidation state of oxygen is $ - 2$ )
$ \Rightarrow x = 7$
So, the oxidation state of chlorine in $HCl{O_4}$ is $7$ .
Chlorine cannot lose more electrons in such cases. So, it cannot be oxidized.
So, $HClO$ and $HCl{O_4}$ are the two substances among them which may be reduced as well as oxidized.
Hence, option B and C are correct.
Note:
Redox reactions are such types of reactions where oxidation and reduction both take place simultaneously. In redox reactions oxidation states of atoms get changed. In this reaction one species undergoes oxidation and on the other hand other species undergoes reduction. The oxidation half is identified by using the oxidation state such as increase in oxidation state or loss of electron is oxidation and the reduction half is identified where oxidation state decreases and gain of electrons takes place.
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