Question

In the chemical reaction,
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+x{{H}_{2}}S{{O}_{4}}+yS{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+z{{H}_{2}}O$
x, y and z are, respectively:
(A) 1, 3, 1
(B) 4, 1, 4
(C) 3, 2, 3
(D) 2, 1, 2

Answer 1

Hint: By a rough observation we can say that the above given reaction is a type of redox reaction which needs to be balanced by some proper stoichiometric coefficients. Balancing the given reaction requires the hit and trial method before we reach the actual stoichiometric values.

Complete Solution :
Let us know some basic steps before we balance the given reaction;
Balancing the reaction-
This is the basic and primary step while evaluating the reactions while we analyse the same on the basis of stoichiometry. We change the stoichiometric coefficients to balance the given reactions instead of changing the subscripts of the given compounds.
Steps to balance redox reaction-
1. We write the oxidation and reduction half reactions for the species that are being reduced or oxidised in the reaction.
2. We then multiply those half-reactions by a suitable number so that we result in having an equal number of electrons on either side.
3. Finally, we add those two equations to cancel out the electrons and hence, obtaining a balanced equation.

Taking these points into consideration, let us solve the given illustration:
Step I-
$S{{O}_{2}}+2{{H}_{2}}O\to {{\left( S{{O}_{4}} \right)}^{2-}}+4{{H}^{+}}+2{{e}^{-}}$
${{\left( C{{r}_{2}}{{O}_{7}} \right)}^{2-}}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O$

Step II-
Multiplying by 3 and adding to (1.2) , we get;
$3S{{O}_{2}}+{{\left( C{{r}_{2}}{{O}_{7}} \right)}^{2-}}+2{{H}^{+}}\to 3{{\left( S{{O}_{4}} \right)}^{2-}}+2C{{r}^{3+}}+{{H}_{2}}O$

Step III-
Now, completing the overall reaction, we get:
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}+3S{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+{{H}_{2}}O$
Therefore, we can see:
x = 1, y = 3, z = 1.
So, the correct answer is “Option A”.

Note: Do note that we have another method too to evaluate the stoichiometric coefficients. This method calls up for balancing each element separately i.e. the total number of any atom on the reactant side will be equal to the total number of the same atom on the product side.
- Solving this for every atom to be balanced; we get sets of linear equations which eventually led to the required stoichiometric coefficients.