Question

Find the element that is oxidized and the one that is reduced 4 NaClO + H2S 4 NaCl + H2SO4

Answer 1

Oxidation is loss of electrons or increase in oxidatition number whilst reduction is the gain of electrons or a decrease in oxidation number.

$4NaClO+H_2S\rightarrow4NaCl+H_2SO_4$

We will find the oxidation state of the chlorine in NaClO:

$\begin{gathered} Na:+1 \\ O:-2 \\ Cl:1+x-2=0 \\ Cl:x-1=0 \\ Cl:x=+1 \end{gathered}$

So chlorine as an oxidation state of +1 in NaClO and -1 in NaCl. There is a decrease in oxidation number so the chlorine is being reduced.

We will determine the oxidation number of S in H2SO4:

$\begin{gathered} H:1 \\ O:-2 \\ S:(2\times1)+(-2\times4)+x=0 \\ S:2-8+x=0 \\ S:-6+x=0 \\ S:x=+6 \end{gathered}$

Sulfur as an oxidation number of -2 in H2S and +6 in H2SO4. There is an increase in the oxidation number so silfur is being oxidized.