2H2O (l) + 2F2(g) → 4HF(aq) + O2(g). In the reaction above, the substance that is being reduced is

To determine which substance is being reduced, we need to look at the changes in the oxidation states of the elements involved in the reaction.

1. Analyze the concept

• Reduction is defined as a decrease in the oxidation state of an element (or the gain of electrons).
• Oxidation is defined as an increase in the oxidation state of an element (or the loss of electrons).
• Oxidation State Rules: • Elements in their standard state (like $F_2$ and $O_2$) have an oxidation state of 0. • In compounds, Hydrogen ($H$) is usually +1. • In compounds, Oxygen ($O$) is usually -2. • In compounds, Fluorine ($F$) is always -1.

2. Step-by-step solution

Let's assign oxidation states to each element in the reaction: $2H_2O(l) + 2F_2(g) \rightarrow 4HF(aq) + O_2(g)$

• Reactants: • In $H_2O$: $H = +1$, $O = -2$ • In $F_2$: $F = 0$ (elemental state)
• Products: • In $HF$: $H = +1$, $F = -1$ • In $O_2$: $O = 0$ (elemental state)

Now, observe the changes:

• Fluorine ($F$): Changes from 0 (in $F_2$) to -1 (in $HF$). Since the oxidation state decreased, Fluorine is reduced.
• Oxygen ($O$): Changes from -2 (in $H_2O$) to 0 (in $O_2$). Since the oxidation state increased, Oxygen is oxidized.

Since the question asks for the substance being reduced, we look for the reactant that contains the element whose oxidation state decreased. That element is Fluorine, which is found in $F_2$.

3. Conclusion

The substance being reduced is $F_2(g)$ because its oxidation state drops from 0 to -1.

Exam Tip

Remember the acronym OIL RIG: Oxidation Is Loss (of electrons/increase in oxidation number), Reduction Is Gain (of electrons/decrease in oxidation number). Always check the elemental forms first—they are the easiest to spot because their oxidation state is always zero!