# Question #559dc

Jun 7, 2017

See explanation.

#### Explanation:

When balancing reactions with charges, you have two things to keep in mind: conservation of charge and conservation of matter. The sum of the charges of the reactants and products must be the same, and the sum of the atoms of each element must be the same before and after the reaction.

One helpful method to balance a reaction with charges is the "half-reaction method." Basically, you split your original equation into two "half-reactions" in which every atom (excluding hydrogen and oxygen) are present both before and after the reaction. In your case the two half reactions would be:
1. ${W}^{6 +} \to {W}^{2 +}$ and
2. $T {l}^{+} \to T {l}^{3 +}$
The first is the reduction half-reaction, because atoms of tungsten (6+) gain electrons, while the second is the oxidation half-reaction, because atoms of thallium (+) lose electrons.

The next step is to balance the charges present before and after the half-reaction. Let's start with the first half-reaction. According to the chemical equation, 4 moles of electrons are gained for each mole of tungsten (6+) that reacts. However, the left side of the half-reaction has a charge of 6+ and the right side has a charge of 2+, and in the half-reaction the charges before and after must be equal, so we will add 4 electrons to the left side, resulting in the following equation:
$4 {e}^{-} + {W}^{6 +} \to {W}^{2 +}$.

Do the same to the other half-reaction and obtain the following:
$T {l}^{+} \to T {l}^{3 +} + 2 {e}^{-}$

We are going to combine the equations, but the electrons must first cancel out because we do not typically show electrons in the balanced chemical equation. So multiply the coefficients of each species in the second half-reaction by 2 and obtain the following:
$2 T {l}^{+} \to 2 T {l}^{3 +} + 4 {e}^{-}$.

We can now combine the two equations:
$2 T {l}^{+} + 4 {e}^{-} + {W}^{6 +} \to 2 T {l}^{3 +} + 4 {e}^{-} \left(+\right) {W}^{2 +}$.
And we can cancel out the $4 {e}^{-}$ and obtain our final balanced chemical equation:
$2 T {l}^{+} + {W}^{6 +} \to 2 T {l}^{3 +} + {W}^{2 +}$.

The net charge is the same before and after the reaction (8+) and there are equal atoms of tungston (2) and thallium (1) present before and after.