72.5k views
5 votes
Find the element that is oxidized and the one that is reduced 4 NaClO + H2S 4 NaCl + H2SO4

User Misman
by
8.0k points

1 Answer

6 votes

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*1)+(-2*4)+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.

User Etienne Tonnelier
by
9.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.