Step-by-step explanation:
Sulfur Reduction Balancing
QUESTION
You are given the equation S + NaOH Na₂S₂O3 + Na₂S +H₂O
i. Explain with electronic concept which substance is oxidised and which is reduced?
ii.Balance the equation by Ion electron or oxidation number method.
iii. Indicate the number of NaOH molecules used.
ANSWER
i. In this reaction, the sodium thiosulfate (Na₂S₂O₃) is being formed from the reaction between sodium hydroxide (NaOH) and sulfur (S). Since sodium thiosulfate has a lower oxidation state of sulfur than elemental sulfur, it can be inferred that sulfur has been reduced, meaning it has gained electrons. On the other hand, sodium hydroxide has not undergone any change in oxidation state, so it can be concluded that it is not oxidized or reduced in this reaction.
ii. To balance the equation using the oxidation number method, we first need to assign oxidation states to each element.
S + NaOH → Na₂S₂O₃ + Na₂S + H₂O
S: 0 → -2
Na: +1 → +1
O: -2 → -2
H: +1 → +1
Now we can see that sulfur has been reduced from 0 to -2, and the oxygen in thiosulfate (S₂O₃²⁻) has been oxidized from -2 to -1. We can balance the equation by adding electrons to the left-hand side to balance the reduction of sulfur and removing electrons from the right-hand side to balance the oxidation of oxygen.
S + 2NaOH + 2e⁻ → Na₂S + 2e⁻ + H₂O
S + 2NaOH → Na₂S + H₂O
S + 2NaOH → Na₂S₂O₃ + Na₂S + H₂O
2S + 4NaOH → Na₂S₂O₃ + 2Na₂S + 2H₂O
Therefore, the balanced equation is: 2S + 4NaOH → Na₂S₂O₃ + 2Na₂S + 2H₂O
iii. The equation indicates that 4 molecules of NaOH are used in the reaction.