Question

Which of the following is a non – redox process?
A.$S{O_4}^{2 - } \to S{O_3}$
B.$C{r_2}{O_7}^{2 - } \to Cr{O_4}^{2 - }$
C.$P{O_4}^{3 - } \to {P_2}{O_7}^{4 - }$
D.${C_2}{O_4}^{2 - } \to C{O_2}$

Answer 1

We need to calculate the oxidation states of atoms in all the options to find out which reaction is redox and which is not.

Complete step by step answer:
Definition –
Redox reaction – A chemical reaction in which there is an increase in oxidation number of one atom and decrease in oxidation number of another atom (both atoms could be different atoms of the same element also) is called a redox reaction.
Non – redox reactions – The reactions in which there is neither decrease nor increase in the oxidation states of any atom are called non – redox reactions.
Oxidation number – It is defined as the total number of electrons lost or gained by an atom in order to form a chemical bond.
Now, we need to calculate the oxidation numbers of each atom in each option in order to determine which reaction is a redox reaction and which reaction is a non – redox reaction.
Considering option A –
Let the oxidation number of $S$ in $S{O_4}^{2 - }$ be $x$. So,
$\Rightarrow x + ( - 2)4 = - 2 \\\ \Rightarrow x - 8 = - 2 \\\ \Rightarrow x = - 2 + 8 \\\ \Rightarrow x = 6 \\\$
Let the oxidation number of $S$ in $S{O_3}$ be $y$. So,
$\Rightarrow y + ( - 2)3 = 0 \\\ \Rightarrow y - 6 = 0 \\\ \Rightarrow y = 6 \\\$
Now, as we can see that the oxidation states of $S$in $S{O_4}^{2 - }$and $S{O_3}$are same, so the reaction is a non –redox reaction.
Considering option B –
Let the oxidation state of $Cr$ in $C{r_2}{O_7}^{2 - }$ be $p$. So,
$\Rightarrow 2p + ( - 2)7 = - 2 \\\ \Rightarrow 2p - 14 = - 2 \\\ \Rightarrow 2p = - 2 + 14 \\\ \Rightarrow 2p = 12 \\\ \Rightarrow p = \frac{{12}}{2} \\\ \Rightarrow p = 6 \\\$
Let the oxidation state of $Cr$ in $Cr{O_4}^{2 - }$ be $q$. So,
$\Rightarrow q + ( - 2)4 = - 2 \\\ \Rightarrow q - 8 = - 2 \\\ \Rightarrow q = - 2 + 8 \\\ \Rightarrow q = 6 \\\$
Now, as we can see that the oxidation states of $Cr$in $C{r_2}{O_7}^{2 - }$ and $Cr{O_4}^{2 - }$ are same, so the reaction is a non –redox reaction.
Considering option C –
Let the oxidation state of $P$ in $P{O_4}^{3 - }$ be $u$. So,
$\Rightarrow u + ( - 2)4 = - 3 \\\ \Rightarrow u - 8 = - 3 \\\ \Rightarrow u = - 3 + 8 \\\ \Rightarrow u = 5 \\\$
Let the oxidation state of $P$ in ${P_2}{O_7}^{4 - }$ be $v$. So,
$\Rightarrow 2v + ( - 2)7 = - 4 \\\ \Rightarrow 2v - 14 = - 4 \\\ \Rightarrow 2v = - 4 + 14 \\\ \Rightarrow 2v = 10 \\\ \Rightarrow v = \frac{{10}}{2} \\\ \Rightarrow v = 5 \\\$
Now, as we can see that the oxidation states of $P$in $P{O_4}^{3 - }$ and ${P_2}{O_7}^{4 - }$ are same, so the reaction is a non –redox reaction.
Considering option D –
Let the oxidation state of $C$ in ${C_2}{O_4}^{2 - }$ be $s$. So,
$\Rightarrow 2s + ( - 2)4 = - 2 \\\ \Rightarrow 2s - 8 = - 2 \\\ \Rightarrow 2s = - 2 + 8 \\\ \Rightarrow 2s = 6 \\\ \Rightarrow s = \frac{6}{2} \\\ \Rightarrow s = 3 \\\$
Let the oxidation state of $C$ in $C{O_2}$ be $t$. So,
$\Rightarrow t + ( - 2)2 = 0 \\\ \Rightarrow t - 4 = 0 \\\ \Rightarrow t = 4 \\\$
As we can see that the oxidation states of $C$ in ${C_2}{O_4}^{2 - }$ and $C{O_2}$ are different, so, this reaction is a redox reaction.

Hence option A,B and C are correct.

Note:
The oxidation state of oxygen is -2 in all compounds.In elemental form, all atoms are at zero oxidation state.