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Question: Balance the following chemical equation using the method of partial equations: \[{K_2}C{r_2}{O_7} ...

Balance the following chemical equation using the method of partial equations:
K2Cr2O7+H2SO4+SO2→K2SO4+Cr2(SO4)3+H2O{K_2}C{r_2}{O_7} + {H_2}S{O_4} + S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O

Explanation

Solution

The balancing of a chemical equation is the process by which the number of atoms of each element are made equal on the reactant and the product by putting stoichiometric coefficients in front of the formula of different compounds and elements.

Complete answer:
The method of partial equations is used when the hit and trial method fails to balance an equation. In this method, the entire equation is fragmented into simpler equations that can be easily balanced by hit and trial methods and then these equations are recombined along with their stoichiometric coefficients to give a net balanced equation.
The given chemical equation is :
K2Cr2O7+H2SO4+SO2→K2SO4+Cr2(SO4)3+H2O{K_2}C{r_2}{O_7} + {H_2}S{O_4} + S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O
We, first start by ignoring the spectator ions Such as potassium ions, hydrogen ions and
sulphate ion. The given reaction can be broken down into two oxidation reduction reactions written as follows:
Cr2O72−→reductionCr3+C{r_2}{O_7}^{2 - }\xrightarrow{{reduction}}C{r^{3 + }}
SO2→oxidationSO42−S{O_2}\xrightarrow{{oxidation}}S{O_4}^{2 - }
We must balance these partial equations by making the elements equal on both sides and using water molecules to balance the oxygen atoms on both sides:
Cr2O72−→reduction2Cr3++7H2OC{r_2}{O_7}^{2 - }\xrightarrow{{reduction}}2C{r^{3 + }} + 7{H_2}O
SO2+2H2O→oxidationSO42−S{O_2} + 2{H_2}O\xrightarrow{{oxidation}}S{O_4}^{2 - }
The hydrogen atoms can now be balanced by using the hydrogen ions that we ignored earlier as spectator ions:
Cr2O72−+14H+→reduction2Cr3++7H2OC{r_2}{O_7}^{2 - } + 14{H^ + }\xrightarrow{{reduction}}2C{r^{3 + }} + 7{H_2}O
SO2+2H2O→oxidationSO42−+4H+S{O_2} + 2{H_2}O\xrightarrow{{oxidation}}S{O_4}^{2 - } + 4{H^ + }
Now, the charges on the sides need to be balanced by adding electrons :
Cr2O72−+14H++6e−→reduction2Cr3++7H2OC{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - }\xrightarrow{{reduction}}2C{r^{3 + }} + 7{H_2}O
SO2+2H2O→oxidationSO42−+4H++2e−S{O_2} + 2{H_2}O\xrightarrow{{oxidation}}S{O_4}^{2 - } + 4{H^ + } + 2{e^ - }
Multiply the second half equation by three so as to balance having an equal number of electrons being involved in the redox reaction.
3SO2+6H2O→oxidation3SO42−+12H++6e−3S{O_2} + 6{H_2}O\xrightarrow{{oxidation}}3S{O_4}^{2 - } + 12{H^ + } + 6{e^ - }
Thus, the electrons cancel out when the two half reactions are added.
Add the two equations and introduce the missing potassium ions and one sulphate ion, and combine the ions wherever required:
K2Cr2O7+H2SO4+3SO2→K2SO4+Cr2(SO4)3+H2O{K_2}C{r_2}{O_7} + {H_2}S{O_4} + 3S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O
⇒\Rightarrow Thus, the balanced chemical equation is :
K2Cr2O7+H2SO4+3SO2→K2SO4+Cr2(SO4)3+H2O{K_2}C{r_2}{O_7} + {H_2}S{O_4} + 3S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O

Note:
Reactions in chemistry follow a certain set of rules known as the laws of chemical composition. One such law is the law of conservation of energy which tells us that mass can neither be conserved nor destroyed i.e. the total mass of the reactants must always be equal to the total mass of products. This rule is used to balance the chemical equations.