Question

# How to write the balanced equation for the reaction between PbS and H${_2}$O${_2}$?

Solution

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In this case we have reduction-oxidation (redox) reaction, because the electrons in the reaction are being exchanged.

$\mathrm{PbS(s)\ +\ H_2O_2(aq) \rightarrow PbSO_4(s)\ +\ H_2O(l)}$

In H${_2}$O${_2}$ oxygen has 1- charge, but in H${_2}$O and PbSO${_4}$ oxygen has 2- charge.

In PbS sulfur has 2- charge, but in PbSO${_4}$ sulfur has 6+ charge.

The balanced equation will be as following:

$\mathrm{PbS(s)\ +\ 4H_2O_2(aq) \rightarrow PbSO_4(s)\ +\ 4H_2O(l)}$

We put the coefficients in front of every compound to make sure that the number of atoms on the left side is equal to the number of atoms on the right side of the equation.