Question

How do you write a balanced overall reaction given the following unbalanced half reactions?
$\begin{aligned} & Li\to L{{i}^{+}} \\\ & B{{a}^{+2}}\to Ba \\\ \end{aligned}$

Answer 1

To balance the reaction, multiply one equation with a constant so that number of electrons lost is equal to the number of electrons gained, then add the two equations, to get the final balanced equation.

Complete step by step answer:
- In order to answer our question, we need to learn about the balancing of chemical equations. We know that matter can neither be created nor be destroyed in a chemical reaction. Moreover, no new species of some different element can suddenly be generated in any reaction. For example, in the reaction $2{{H}_{2}}+{{O}_{2}}\to 2{{H}_{2}}O$, we can see that both the number of atoms of hydrogen and oxygen in the reactant side and product side of the reaction are same. There are 4 hydrogen atoms and 2 oxygen atoms throughout the reaction. Moreover, when hydrogen and oxygen are mixed, the end product is water, which consists of hydrogen and oxygen only. No new element like nitrogen or argon has interfered in the reaction.
- However, when it comes to the balancing of ionic reactions, there is a little bit of a difference. In ionic reactions, there is exchange of electrons, and electrons are exchanged because out of protons, neutrons and electrons, electrons are the only mobile. Even in this case, the mass conservation principle is applied. So, we have to balance them in such a way that the number of electrons transferred is numerically equal to the number of electrons gained. Now, we have:

& Li\to L{{i}^{+}}+{{e}^{-}}...........(1) \\\ & B{{a}^{2+}}+2{{e}^{-}}\to Ba......(2) \\\ \end{aligned}$$ Clearly, there is transfer of 2 electrons, so, we multiply equation (1) by 2 and we have: $$2Li\to 2L{{i}^{+}}+2{{e}^{-}}...........(3)$$ Now, by adding equations (2) and (3), we can get the final balanced equation that is: $$2Li+B{{a}^{2+}}\to 2L{{i}^{+}}+Ba$$ Hence, we get the required balanced equation. **Note:** In the case of ionic reactions, the charge is treated as atomicity, which cannot be changed for any reaction. However, the coefficients can be changed by multiplying with a constant.