Animated Solution for Chemistry - Redox Reactions: $2{\text{MnO}}_4^{-}+b{\text{C}}_2{\text{O}}_4^{2-}+c{\text{H}}^{+}⟶x{\text{Mn}}^{2+}+y{\text{CO}}_2+z{\text{H}}_2\text{O}$ If the above equation is balanced with integer coefficients, the value of $c$ is ......... .

The process of balancing a complex redox reaction can often feel like untangling a massive knot. But fear not! By breaking it down systematically using the ion-electron method (also known as the half-reaction method), we can turn this daunting task into a smooth, logical sequence of steps. Let's dive into the fascinating dance of electrons between permanganate and oxalate ions.

We are given an unbalanced chemical equation:

Our ultimate goal is to find the exact integer value for the coefficient $c$, which represents the number of hydrogen ions (${\text{H}}^{+}$) required to balance the reaction. The presence of ${\text{H}}^{+}$ immediately tells us that this reaction is taking place in an acidic medium. This is a crucial piece of information because it dictates how we will balance the oxygen and hydrogen atoms later on.

The very first step in our journey is to divide and conquer. We split the main, complex equation into two simpler half-reactions: one dedicated entirely to oxidation, and the other to reduction.

Looking at the species involved, we can see that the oxalate ion (${\text{C}}_2{\text{O}}_4^{2-}$) is converting into carbon dioxide (${\text{CO}}_2$). In this process, the oxidation state of carbon increases from $+3$ to $+4$. This is our oxidation half-reaction:

Simultaneously, the intensely purple permanganate ion (${\text{MnO}}_4^{-}$) is transforming into the nearly colorless manganese(II) ion (${\text{Mn}}^{2+}$). Here, the oxidation state of manganese drops dramatically from $+7$ to $+2$. This is our reduction half-reaction:

Now, we must ensure that the law of conservation of mass is obeyed for every element. We start with atoms other than oxygen and hydrogen.

In the oxidation half, we have two carbon atoms on the reactant side but only one on the product side. We fix this by placing a coefficient of $2$ in front of ${\text{CO}}_2$:

The manganese atoms in the reduction half are already balanced, with one on each side.

Next, we tackle the oxygen and hydrogen atoms. The oxidation half now has four oxygen atoms on both sides, so it is perfectly balanced! However, the reduction half has four oxygen atoms on the left and none on the right. To balance the oxygens, we add four water molecules (${\text{H}}_2\text{O}$) to the right side:

Adding water introduced eight hydrogen atoms to the right side. Because we are in an acidic medium, we balance these by adding eight ${\text{H}}^{+}$ ions to the left side:

With the atoms balanced, we must now balance the electrical charges by adding electrons ($e^{-}$). This is where many students make a silly mistake, so we must be careful!

In the oxidation half, the left side has a net charge of $-2$, while the right side is neutral ($0$). To balance this, we add two electrons to the right side:

In the reduction half, the left side has eight positive charges and one negative charge, resulting in a net charge of $+7$. The right side has a net charge of $+2$. To bring $+7$ down to $+2$, we must add five electrons to the left side:

In any valid redox reaction, the number of electrons lost during oxidation must exactly equal the number of electrons gained during reduction. We cannot have free electrons floating around in our final equation!

Currently, our oxidation half produces $2$ electrons, while our reduction half consumes $5$ electrons. To equalize them, we find the lowest common multiple, which is $10$. We multiply the entire oxidation half-reaction by $5$, and the entire reduction half-reaction by $2$:

Oxidation (multiplied by 5):

Reduction (multiplied by 2):

Now for the most satisfying part! We add the two multiplied half-reactions together. The $10$ electrons on both sides perfectly cancel each other out.

Combining the remaining terms, we arrive at our completely balanced overall redox equation:

Finally, we return to the specific question asked. We needed to find the value of $c$, which is the stoichiometric coefficient of the ${\text{H}}^{+}$ ions. By comparing our beautifully balanced equation with the general form provided in the problem, we can clearly see that the coefficient of ${\text{H}}^{+}$ is $16$.

Therefore, our final answer is $16$.