Question

In balancing the half reaction ${S_2}{O_3}^{2 - } \to S\left( s \right)$, the number of electrons that must be added is:
A.2 on the right
B.2 on the left
C.3 on the right
D.4 on the left

Answer 1

Hint: We have to balance the chemical equation as well the chemical charges on both the sides to determine the number of electrons which should be added.

Complete step by step answer:
The chemical equation is given as,
${S_2}{O_3}^{2 - } \to S\left( s \right)$
We can see that the equation remains unbalanced.
We need two moles of sulfur on the product side and to balance oxygen, we add water. To balance hydrogen we add ${H^ + }$.
We can write the balanced reaction as,
$6{H^ + } + {S_2}{O_3}^{2 - } \to 2S + 3{H_2}O$
Let us now balance the charge,
In the left hand side, we have +6 and -2 charge, so difference is +4 charge $\left( { + 6 - 2 = + 4} \right)$
On the right hand side, the charge is zero.
We know that charge should be equal on both sides, which means on the left hand side the charge should be zero.
We will get zero charge on the left hand side when we add four electrons on the left hand side.
$4{e^ - } + 6{H^ + } + {S_2}{O_3}^{2 - } \to 2S + 3{H_2}O$

So, the correct answer is Option D.

Note:
We can also determine the number of electrons added using the oxidation state. The oxidation state of sulfur in ${S_2}{O_3}^{2 - }$ is +2 and the oxidation state of elemental sulfur is zero.
$\mathop {{S_2}}\limits_{2 + } {O_3}^{2 - } \to \mathop S\limits_0 \left( s \right)$
As the change in oxidation state of each sulfur atom is +2.
$4{e^ - } + {S_2}^{2 + } \to 2{S^0}$
Therefore, the number of electrons to be added is four on the left.
We can define redox reactions as chemical reactions in which transfer of electrons takes places between two reactants. We can find the transfer of electrons by observing the change in oxidation states of the species that are reacting. The redox reaction takes place in batteries (or) electrochemical cells.