Question

Find the oxidation number of elements in each case C in$\;C{H_4}$, ${C_2}{H_6}$, ${C_3}{H_8}$, ${C_2}{H_2}$,${H_2}{C_2}{O_4}$, and $C{O_2}$
A)$- 4$,$- 3$,$- \dfrac{8}{3}$,$- 2$,$- 1$,$+ 3$,$+ 4$
B)$- 3$,$- 3$,$- \dfrac{8}{3}$,$- 2$,$- 2$,$+ 3$,$+ 4$
C)$- 2$,$- 3$,$- 8$,$- 2$,$- 1$,$+ 3$,$+ 4$
D) None of the above.

Answer 1

We know that electrochemical equivalent of an element is defined as the mass of an element deposited at an electrode when one ampere of electricity is flows through an electrolytic cell for one second and it is mentioned as$Z$.

Complete step by step answer:
The oxidation number of carbon atom in carbon containing compounds can be calculated using the formula $\dfrac{{2{n_o} - {n_H}}}{{{n_c}}}$
Where$2{n_o}$, ${n_H}$, ${n_C}$ represents the number of oxygen, hydrogen and carbon atoms respectively.
For methane, the oxidation number of carbon atom$= \dfrac{{0 - 4}}{1} = - 4$
For ${C_2}{H_6}$, the oxidation number of carbon atom$= \dfrac{{0 - 6}}{2} = - 3$
For ${C_3}{H_8}$, the oxidation number of carbon atom$= \dfrac{{0 - 8}}{3} = - \dfrac{8}{3}$
For ${C_2}{H_4}$, the oxidation number of carbon atom$= \dfrac{{0 - 4}}{2} = - 2$
For ${H_2}{C_2}{O_4}$, the oxidation number of carbon atom$= \dfrac{{2 \times 4 - 4}}{2} = + 3$
For $C{O_2}$, the oxidation number of carbon atom$= \dfrac{{2 \times 2 - 0}}{1} = + 4$
Thus the oxidation number of carbon is $\;C{H_4}$,${C_2}{H_6}$,${C_3}{H_8}$,${C_2}{H_2}$,${H_2}{C_2}{O_4}$, and $C{O_2}$ is $- 4$,$- 3$,$- \dfrac{8}{3}$,$- 2$,$- 1$,$+ 3$,$+ 4$respectively.

Therefore, the option A is correct.

Additional note:
Let us see few rules for oxidation numbers,
A free element will be zero as its oxidation number.
Monatomic ions will have an oxidation number equal to charge of the ion.
In hydrogen, the oxidation number is ${\text{ + 1}}$, when combined with elements having less electronegativity; the oxidation number of hydrogen is -1.
In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.
Group 1 elements will have +1 oxidation number.
Group 2 elements will have +2 oxidation numbers.
Group 17 elements will have -1 oxidation number.
Sum of oxidation numbers of all atoms in neutral compounds is zero.

Note:
As we know that the reduction is an opposite process of oxidation. If there is gain of electrons by a molecule, atom or ion then it is called as reduction it occurs when the oxidation state of the species is decreased. Reducing agent is the one which is oxidized in a reaction. Reducing agent reduces the oxidizing agent.
Consider the reaction,
$Pb\left( s \right) + Pb{O_2}\left( s \right) + 2{H_2}S{O_4}\left( {aq} \right)\xrightarrow{{}}2PbS{O_4}\left( s \right) + 2{H_2}O\left( l \right)$
In the above reaction, the oxidation state of lead$\left( {Pb} \right)$ is zero and the oxidation state of lead in $\left( {Pb{O_2}} \right)$ is $+ 4$ and the oxidation state of lead in $\left( {PbS{O_4}} \right)$is$+ 2$. Hence, the oxidized species is lead and the species which is reduced is lead oxide. The oxidizing agent is lead sulfate and the reducing agent is lead.