Question

Divide the following redox equation into oxidation half equation and reduction half equation:
$P\left( s \right) + P{O_4}^{3 - }\left( {aq} \right) \to HP{O_3}^{2 - }\left( {aq} \right)$

Answer 1

Hint : A redox reaction can be defined as a reaction in which there is transfer of electrons between two species. Redox reactions are very common reactions and widely used reactions. They are also vital to some basic functions of life.

Complete Step By Step Answer:
Oxidation involves loose or removal of electrons during oxidation there is increase in oxidation state. Reducing agent is the one which reduces others and oxidises itself.
$M \to {M^{2 + }} + 2e$
Reduction involves gain of electrons, there is decrease in oxidation state after reduction. Oxidizing agents oxidize others and itself gets reduced.
${M^{2 + }} + 2e \to M$
Now, coming to our question we have to divide the redox equation into oxidation and reduction half equations.
$P\left( s \right) + P{O_4}^{3 - }\left( {aq} \right) \to HP{O_3}^{2 - }\left( {aq} \right)$
Oxidation half equation: Oxidation state should increase
$P\left( s \right) \to HP{O_3}^{2 - }\left( {aq} \right)$
$x = 0$ $x + 1 + 3\left( { - 2} \right) = - 2$
$x = 3$
Oxidation state increases from $0$ to 3. Thus, the oxidation reaction is there.
Reduction half equation: Oxidation state should decrease.
$P{O_4}^{3 - }\left( {aq} \right) \to HP{O_3}^{2 - }$
$x + 4\left( { - 2} \right) = - 3$ $x = 3$
$x = 5$
Oxidation state decreases from $5$ to $3$ . Thus, a reduction reaction is there.
Thus, the redox equation is divided into oxidation half equation and reduction half equation.

Additional Information:
Redox reactions have many diverse applications. For example: redox reaction in electrochemistry, redox reaction in combustion, redox reaction in photosynthesis, Qualitative analysis. Electrochemical cells have a lot of applications in the industrial sector as well and electrolysis is used in many production processes which are based on redox reactions.

Note :
In such types of redox equations, we just have to check the oxidation states in order to find out whether oxidation is taking place or reduction. One should know how to find the oxidation state in order to check the oxidation or reduction of half cells.