Question

Three solutions each of 100 mL containing 0.4 M $\text{A}{{\text{s}}_{\text{2}}}{{\text{S}}_{\text{3}}}$, 5M NaOH and 6M ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ respectively was mixed to form $\text{AsO}_{\text{4}}^{3-}$and $\text{SO}_{\text{4}}^{\text{2}-}$as products. Sum of coefficients of reactants and products in a balanced equation is:

Answer 1

Hint: We should first write the balanced equation and then write the oxidation and reduction reaction separately. Then we can calculate the coefficient.

Complete step by step solution:
So, from the above question we will first write out the reaction.
$A{{s}_{\text{2}}}{{S}_{\text{3}}}+NaOH+{{H}_{\text{2}}}{{O}_{\text{2}}}\to AsO_{\text{4}}^{\text{3}-}+{{H}_{\text{2}}}O+SO_{4}^{\text{2}-}$
The reaction present in question is represented above.
Let us now break the above reaction. Breaking the above reaction means, we have to distribute it into the oxidation and reduction part. And also we should now revise about oxidation and reduction concepts to correctly write the separate reaction.
Oxidation: We should know that oxidation is the loss of electrons or we can say it as an increase in the oxidation state of an atom, an ion, or of certain atoms in a molecule.
Reduction: We should know that reduction is the gain of electrons or we can say it as a decrease in the oxidation state of an atom, an ion, or of certain atoms in a molecule.
Let us now write the reaction.

First we will write the reaction of oxidation.

$\text{As}_{\text{2}}^{3+}+\text{16OH}\to \text{2AsO}_{\text{4}}^{3-}+\text{8}{{\text{H}}_{\text{2}}}\text{O}+\text{4}{{\text{e}}^{-}}~~~$

The above is the oxidation reaction of As. Its oxidation state increases from +3 to +5. We balance the above reaction.

$\text{S}_{\text{3}}^{2-}+\text{24OH}\to \text{3SO}_{\text{4}}^{\text{3-}}+\text{12}{{\text{H}}_{\text{2}}}\text{O}+\text{24}{{\text{e}}^{-}}$

The above is the oxidation reaction of sulphur. Its oxidation state increases from -2 to +5. We should also balance the above reaction.
Now, we will write the combined form of both the reactions.

$\text{A}{{\text{s}}_{\text{2}}}{{\text{S}}_{\text{3}}}+\text{4}0\text{OH}\to \text{2AsO}_{\text{4}}^{3-}+\text{3SO}_{\text{4}}^{\text{2}-}+\text{2}0{{\text{H}}_{\text{2}}}\text{O}+\text{28}{{\text{e}}^{-}}$ Equation (1)

Now, we will write the reduction part of the main reaction.

${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}+\text{2}{{\text{e}}^{-}}\to \text{2OH}$ Equation (2)

Now, we will multiply reaction 2 by 14 and we will add reaction 2 and 3 to get the sum of coefficients.

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{A}{{\text{s}}_{\text{2}}}{{\text{S}}_{\text{3}}}+12\text{OH+14}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\to \text{2AsO}_{\text{4}}^{3-}+\text{3SO}_{\text{4}}^{\text{2}-}+\text{2}0{{\text{H}}_{\text{2}}}\text{O} \\\ & Coefficient\to \,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,20\,\,\,\,\,\,\,\, \\\ \end{aligned}$$ If we add the coefficient then the sum is 52. Hence, the answer is 52. Note: We should know that the original view of oxidation and reduction is that of adding or removing oxygen. An alternative approach that we can use to describe oxidation as the loss of hydrogen and reduction as the gaining of hydrogen