Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)
(a) CH4(g) + S(s) �¨ CS2(l) + H2S(g)
To balance the given reaction: CH4(g) + S(s) → CS2(l) + H2S(g), we will go step-by-step.
First, let's balance the carbon (C) atoms:
There is one carbon atom on both sides, so the carbon is already balanced.
Next, let's balance the hydrogen (H) atoms:
On the left side, there are four hydrogen atoms from CH4(g), and only two hydrogen atoms from H2S(g) on the right side. To balance the hydrogen atoms, we need to put a coefficient of 2 in front of H2S(g), giving us:
CH4(g) + S(s) → CS2(l) + 2H2S(g)
Now, let's balance the sulfur (S) atoms:
On the left side, there is one sulfur atom from S(s), and on the right side, there is one sulfur atom from CS2(l) and two sulfur atoms from H2S(g). To balance the sulfur atoms, we need to put a coefficient of 1 in front of S(s), giving us:
CH4(g) + 1S(s) → CS2(l) + 2H2S(g)
Finally, let's check if the reaction is already balanced:
On both sides, there is one carbon atom, four hydrogen atoms, and one sulfur atom. Therefore, the reaction is now balanced.
Identifying the substance reduced and the substance oxidized:
In this reaction, the sulfur (S) in the reactant S(s) is oxidized because it goes from an oxidation state of 0 to -2 in H2S(g). The carbon (C) in the reactant CH4(g) is reduced because it goes from an oxidation state of -4 to +4 in CS2(l).
The balanced reaction is:
CH4(g) + S(s) → CS2(l) + 2H2S(g)
To balance the given reaction by inspection, follow these steps:
Step 1: Identify the oxidation states of each element in the reaction.
CH4(g) + S(s) → CS2(l) + H2S(g)
In CH4, since hydrogen (H) usually has an oxidation state of +1, and methane (CH4) is neutral, the carbon (C) atom must have an oxidation state of -4.
In S(s), sulfur (S) is in its elemental form, which means its oxidation state is 0.
Step 2: Identify the substances being reduced and oxidized.
In the reaction, the carbon (C) in CH4 is oxidized from an oxidation state of -4 to +4 in CS2. Since the oxidation state increases, carbon (C) is being oxidized.
In the reaction, the sulfur (S) in S(s) is reduced from an oxidation state of 0 to -2 in H2S. Since the oxidation state decreases, sulfur (S) is being reduced.
Step 3: Balance the reaction by adjusting the coefficients in front of the reactants and products.
CH4(g) + S(s) → CS2(l) + H2S(g)
At this point, you can observe the number and types of atoms on both sides of the equation and start balancing:
Carbon (C): There is one C atom on both sides, so it is already balanced.
Hydrogen (H): There are four H atoms on the left side and two H atoms on the right side, so you need to balance the H atoms by inserting a coefficient of 2 in front of H2S.
Sulfur (S): There is one S atom on both sides, so it is already balanced.
Now the balanced equation is:
CH4(g) + S(s) → CS2(l) + 2H2S(g)
In this balanced equation, carbon (C) is being oxidized from -4 to +4, and sulfur (S) is being reduced from 0 to -2.