For the oxidation- reduction reaction equation given here,
2Na+S -> Na2S
indicate how many electrons are transferred in the formation of one formula unit of produce.
2
2Na ==> 2Na^+ + 2e
S + 2e ==> S^=
To determine the number of electrons transferred in the formation of one formula unit of product in the oxidation-reduction reaction equation, we need to assign oxidation numbers to each element in the reaction.
In the reaction equation:
2Na + S -> Na2S
The oxidation number of sodium (Na) is +1, and the oxidation number of sulfur (S) is 0 because it is in its elemental form.
In the product, sodium sulfide (Na2S), the oxidation number of sodium (Na) is +1, and the oxidation number of sulfur (S) is -2.
Therefore, each sodium atom loses one electron (from +1 to 0 oxidation state), and sulfur gains two electrons (from 0 to -2 oxidation state).
Since the reaction involves 2 sodium atoms, we can conclude that in the formation of one formula unit of sodium sulfide (Na2S), 2 electrons are transferred.
To determine the number of electrons transferred in a redox reaction, you need to compare the oxidation states of the elements before and after the reaction.
In this reaction, sodium (Na) has an oxidation state of 0 in its elemental form, and sulfur (S) has an oxidation state of 0 as well.
After the reaction, sodium forms Na+ ions with an oxidation state of +1, and sulfur forms S2- ions with an oxidation state of -2.
Now, consider the equation:
2Na + S -> Na2S
In the reactants, there are two sodium atoms and one sulfur atom. Each sodium atom loses one electron to become a Na+ ion, which means a total of 2 electrons are transferred.
Similarly, each sulfur atom gains two electrons to become an S2- ion, which means a total of 2 electrons are transferred.
So, in the formation of one formula unit of Na2S, a total of 2 electrons are transferred.