Questions: How many electrons are transferred in the reaction represented by the following equation? (9.1) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) (a) 0 (b) 2 (c) 4 (d) 8

Solution Steps

First, we need to determine the oxidation states of the elements in the reactants and products.

For \(\mathrm{CH}_4\):

• Carbon (C) in \(\mathrm{CH}_4\) has an oxidation state of -4.
• Hydrogen (H) in \(\mathrm{CH}_4\) has an oxidation state of +1.

For \(\mathrm{O}_2\):

• Oxygen (O) in \(\mathrm{O}_2\) has an oxidation state of 0.

For \(\mathrm{CO}_2\):

• Carbon (C) in \(\mathrm{CO}_2\) has an oxidation state of +4.
• Oxygen (O) in \(\mathrm{CO}_2\) has an oxidation state of -2.

For \(\mathrm{H}_2\mathrm{O}\):

• Hydrogen (H) in \(\mathrm{H}_2\mathrm{O}\) has an oxidation state of +1.
• Oxygen (O) in \(\mathrm{H}_2\mathrm{O}\) has an oxidation state of -2.

Next, we calculate the change in oxidation states for the elements that undergo a change.

For Carbon:

• In \(\mathrm{CH}_4\), the oxidation state of C is -4.
• In \(\mathrm{CO}_2\), the oxidation state of C is +4.
• Change in oxidation state for C: \(+4 - (-4) = +8\).

For Oxygen:

• In \(\mathrm{O}_2\), the oxidation state of O is 0.
• In \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\), the oxidation state of O is -2.
• Each \(\mathrm{O}_2\) molecule (2 atoms) changes from 0 to -2.
• Change in oxidation state for each O atom: \(0 - (-2) = -2\).
• Total change for 2 \(\mathrm{O}_2\) molecules (4 atoms): \(4 \times (-2) = -8\).

The total number of electrons transferred is the sum of the changes in oxidation states for the elements involved.

• Carbon: \(+8\) electrons (oxidation)
• Oxygen: \(-8\) electrons (reduction)

Since the number of electrons lost by carbon is equal to the number of electrons gained by oxygen, the total number of electrons transferred in the reaction is 8.

Final Answer