Question

Balance the following equation by oxidation number method.
Al + KMnO$_4$ +H$_2$SO$_4$ $\rightarrow$ Al$_2$(SO$_4$)$_3$ +K$_2$SO$_4$+ MnSO$_4$ + H$_2O$

Answer 1

Hint: This question is based on the concept of oxidation number method. For this method we must know how to calculate the oxidation number. The reactants, and products are known, so we will assign the oxidation number, and identify whether there is reduction, or oxidation going on. Further, it will lead to the balancing of a chemical equation.

Complete step by step answer:
Now, to balance the chemical equation by oxidation number method, first let us write the unbalanced chemical equation
Al + KMnO$_4$ +H$_2$SO$_4$ $\rightarrow$ Al$_2$(SO$_4$)$_3$ +K$_2$SO$_4$+ MnSO$_4$ + H$_2$O

As we know, this is a redox reaction means oxidation-reduction reaction. Secondly, assign the oxidation number of each atom in the equation.
So, let us do it for the left hand side Al represents 0, Mn represents 7, K represents +1, and O represents -2.
For the right hand side Al represents +3, and Mn represents +2.
So, after assigning the oxidation number, we can say that KMnO$_4$ is an oxidizing agent, and Al acts a reducing agent.
The reaction can be written as:
6Mn$^{VII}$ + 30e$^{-}$ $\rightarrow$ 6Mn$^{II}$ (reduction)
10Al$^{0}$ + 30e$^{-}$ $\rightarrow$ 10Al$^{III}$ (oxidation)
Now, we can write these equations by combining them, and we will balance all the atoms, then the chemical equation is

\[10A{{l}^{0}}+6KM{{n}^{VII}}{{O}_{4}}+24{{H}_{2}}S{{O}_{4}}\to 5A{{l}_{2}}^{III}{{(S{{O}_{4}})}_{3}}+3{{K}_{2}}S{{O}_{4}}+6M{{n}^{II}}S{{O}_{4}}+24{{H}_{2}}O\]

Thus, in the last we can conclude that above mentioned is the balanced equation mainly representing the oxidation number of Mn, and Al.

Note: Don’t get confused while assigning the oxidation number. If you don’t know the oxidation number for each atom, just assign to the main reducing, and oxidizing agent. The oxidising agent means when oxidation number decreases like in case of Mn, and the reducing agent means when the oxidation number increases as shown in Al.