Question

How to balance the following redox problems using both methods?
\[S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+KMn{{O}_{4}}+{{H}_{2}}O\to {{H}_{3}}Sb{{O}_{4}}+{{K}_{2}}S{{O}_{4}}+MnS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\]

Answer 1

Hint: A reaction in which electrons are transferred between two species is known as an oxidation-reduction reaction or a redox reaction. Since an electron is transferred, either by gaining an electron or by losing an electron, the oxidation number of the atom, ion, or molecule also changes.

Complete answer:
The equations in which the coefficients of the reactants and the product give an equal number of atoms for each element is known as a balanced equation.
Redox reactions can be balanced by two methods
1. Oxidation number method: This method uses the difference in the oxidizing agent oxidation number and the reducing agent oxidation number. The steps involved are
- Write the oxidation number of each atom in the equation above the atom.
\[Sb_{2}^{+3}{{({{S}^{+6}}O_{4}^{-2})}_{3}}+{{K}^{+1}}M{{n}^{+7}}O_{4}^{-2}+H_{2}^{+1}{{O}^{-2}}\to H_{3}^{+1}S{{b}^{+5}}O_{4}^{-2}+K_{2}^{+1}{{S}^{+6}}O_{4}^{-2}+M{{n}^{+2}}{{S}^{+6}}O_{4}^{-2}+H_{2}^{+1}{{S}^{+6}}O_{4}^{-2}\]
 - Identify which atoms are oxidized and which are reduced.
Here, the Sb atom is being oxidized as its oxidation number increases from +3 to +5, so the increase in the number of electrons is 2. The atom being reduced is the Mn atom, as its oxidation number decreases to -5 from +2, so the decrease in the number of electrons is 5.
- Now in a balanced redox reaction, the number of electrons gained is equal to the number of electrons lost. Hence, we need to equalize the change in the number of electrons or the oxidation number.
For every 2 atoms of Mn, 5 atoms of Sb are required. Hence for every 4 atoms of Mn, 10 atoms of Sb are required. Now the increase and decrease in the number of electrons are +20 and -20.
- Insert these coefficients in the reaction equation.
\[5S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+4KMn{{O}_{4}}+{{H}_{2}}O\to 10{{H}_{3}}Sb{{O}_{4}}+{{K}_{2}}S{{O}_{4}}+4MnS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\]
- Place required coefficients for molecules that do not participate in the redox process.
For K, \[5S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+4KMn{{O}_{4}}+{{H}_{2}}O\to 10{{H}_{3}}Sb{{O}_{4}}+2{{K}_{2}}S{{O}_{4}}+4MnS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\].
For S, \[5S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+4KMn{{O}_{4}}+{{H}_{2}}O\to 10{{H}_{3}}Sb{{O}_{4}}+2{{K}_{2}}S{{O}_{4}}+4MnS{{O}_{4}}+9{{H}_{2}}S{{O}_{4}}\].
For O, \[5S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+4KMn{{O}_{4}}+24{{H}_{2}}O\to 10{{H}_{3}}Sb{{O}_{4}}+2{{K}_{2}}S{{O}_{4}}+4MnS{{O}_{4}}+9{{H}_{2}}S{{O}_{4}}\].
- Check the balancing for both charge and the number of atoms, and write the balanced equation.
Since all the atoms and the charges are equalized, the balanced equation is
\[5S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+4KMn{{O}_{4}}+24{{H}_{2}}O\to 10{{H}_{3}}Sb{{O}_{4}}+2{{K}_{2}}S{{O}_{4}}+4MnS{{O}_{4}}+9{{H}_{2}}S{{O}_{4}}\]

2. Half-reaction method: This method is based on the division of the redox reactions into the reduction half and the oxidation half. The steps involved are
- We begin by writing the ionic equation of the reaction. This equation is unbalanced.
\[S{{b}^{+3}}+SO_{4}^{-2}+{{K}^{+}}+MnO_{4}^{-}+{{H}_{2}}O\to {{H}^{+}}+SbO_{4}^{-3}+{{K}^{+}}+SO_{4}^{-2}+M{{n}^{+2}}+SO_{4}^{-2}+{{H}^{+}}+SO_{4}^{-2}\]
- Then, every ion that occurs on both sides of the equation is eliminated. Hydrogen ion and water molecules are eliminated as well.
\[S{{b}^{+3}}+MnO_{4}^{-}\to SbO_{4}^{-3}+M{{n}^{+2}}\]
This equation is known as the skeleton ionic equation. From this equation, we can see that \[MnO_{4}^{-}\] is reduced to \[M{{n}^{+2}}\] and \[S{{b}^{+3}}\] is oxidized to \[SbO_{4}^{-3}\].
- Now we divide the reaction into two halves and balance all atoms except hydrogen and oxygen.
\[\begin{align}
  & S{{b}^{+3}}\to SbO_{4}^{-3} \\
 & MnO_{4}^{-}\to M{{n}^{+2}} \\
\end{align}\]
- Now we balance the oxygen atom.
\[\begin{align}
  & S{{b}^{+3}}+4{{H}_{2}}O\to SbO_{4}^{-3} \\
 & MnO_{4}^{-}\to M{{n}^{+2}}+4{{H}_{2}}O \\
\end{align}\]
- Now we balance the hydrogen atom.
\[\begin{align}
  & S{{b}^{+3}}+4{{H}_{2}}O\to SbO_{4}^{-3}+8{{H}^{+}} \\
 & MnO_{4}^{-}+8{{H}^{+}}\to M{{n}^{+2}}+4{{H}_{2}}O \\
\end{align}\]
- The next step is to balance the charge and equalize the electrons transferred.
\[\begin{align}
  & 5\times [S{{b}^{+3}}+4{{H}_{2}}O\to SbO_{4}^{-3}+8{{H}^{+}}+2{{e}^{-}}]=5S{{b}^{+3}}+20{{H}_{2}}O\to 5SbO_{4}^{-3}+40{{H}^{+}}+10{{e}^{-}} \\
 & 2\times [MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{+2}}+4{{H}_{2}}O]=2MnO_{4}^{-}+16{{H}^{+}}+10{{e}^{-}}\to 2M{{n}^{+2}}+8{{H}_{2}}O \\
 & \\
\end{align}\]
- Now we add the two half-reactions and check the balance of charge and mass.
\[5S{{b}^{+3}}+2MnO_{4}^{-}+12{{H}_{2}}O\to 5SbO_{4}^{-3}+2M{{n}^{+2}}+24{{H}^{+}}\]
- Since to match the charge of \[5S{{b}^{+3}}\], we would have to add $7.5SO_{4}^{-2}$, the equation is multiplied by 2 before adding the non-participating ions.
\[10S{{b}^{+3}}+4MnO_{4}^{-}+24{{H}_{2}}O\to 10SbO_{4}^{-3}+4M{{n}^{+2}}+48{{H}^{+}}\]
- Now to the balanced ionic equation, we add the ions which do not participate in the redox reaction.
\[\begin{align}
  & 10S{{b}^{+3}}+4MnO_{4}^{-}+24{{H}_{2}}O\to 10SbO_{4}^{-3}+30{{H}^{+}}+4M{{n}^{+2}}+18{{H}^{+}} \\
 & +15SO_{4}^{-2}+4{{K}^{+}}\text{ }4SO_{4}^{-2}+9SO_{4}^{-2} \\
\end{align}\]
  - The balanced molecular reaction can be written as
\[5S{{b}_{2}}{{(S{{O}_{4}})}_{3}}+4KMn{{O}_{4}}+24{{H}_{2}}O\to 10{{H}_{3}}Sb{{O}_{4}}+2{{K}_{2}}S{{O}_{4}}+4MnS{{O}_{4}}+9{{H}_{2}}S{{O}_{4}}\]

Note:
Even though there is no significant difference between the half-reaction method and the oxidation number method, in a reaction in which the substances are in the aqueous solution, the half-reaction method is preferred over the oxidation number method.