Problem 50 - Chemistry and Chemical Reactivity

Question

In the following reactions, decide which reactant is oxidized and which is reduced. Designate the oxidizing agent and the reducing agent. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+3 \mathrm{Sn}^{2+}(\mathrm{aq})+14 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Sn}^{4+}(\mathrm{aq})+21 \mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{FeS}(\mathrm{s})+3 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{NO}(\mathrm{g})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

(a) \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is the oxidizing agent; \( \mathrm{Sn}^{2+} \) is the reducing agent. (b) \( \mathrm{NO}_{3}^{-} \) is the oxidizing agent; \( \mathrm{FeS} \) is the reducing agent.

Step 1: Identify Oxidation States (Reaction a)

For the reaction \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 3 \mathrm{Sn}^{2+} + 14 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{Sn}^{4+} + 21 \mathrm{H}_{2} \mathrm{O} \), determine the change in oxidation states: - The oxidation state of \( \mathrm{Cr} \) in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is +6. It changes to +3 in \( \mathrm{Cr}^{3+} \), indicating reduction.- The oxidation state of \( \mathrm{Sn} \) changes from +2 in \( \mathrm{Sn}^{2+} \) to +4 in \( \mathrm{Sn}^{4+} \), indicating oxidation.

Step 2: Identify Oxidizing and Reducing Agents (Reaction a)

The oxidizing agent is the substance that gets reduced, and the reducing agent is the one that gets oxidized:- \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is the oxidizing agent as it gains electrons and is reduced.- \( \mathrm{Sn}^{2+} \) is the reducing agent as it loses electrons and is oxidized.

Step 3: Identify Oxidation States (Reaction b)

For the reaction \( \mathrm{FeS} + 3 \mathrm{NO}_{3}^{-} + 4 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow 3 \mathrm{NO} + \mathrm{SO}_{4}^{2-} + \mathrm{Fe}^{3+} + 6 \mathrm{H}_{2} \mathrm{O} \), determine the change in oxidation states:- \( \mathrm{Fe} \) in \( \mathrm{FeS} \) increases from 0 to +3 in \( \mathrm{Fe}^{3+} \), indicating oxidation.- \( \mathrm{N} \) in \( \mathrm{NO}_{3}^{-} \) changes from +5 to +2 in \( \mathrm{NO} \), indicating reduction.

Step 4: Identify Oxidizing and Reducing Agents (Reaction b)

The oxidizing agent is the substance that gets reduced, and the reducing agent is the one that gets oxidized:- \( \mathrm{NO}_{3}^{-} \) is the oxidizing agent as it is reduced.- \( \mathrm{FeS} \) is the reducing agent as it is oxidized.

Key Concepts

Oxidation States

Oxidation states, sometimes known as oxidation numbers, are used to track how electrons are transferred in redox reactions. These numbers help us determine which atoms are gaining or losing electrons. Here is a simple way to understand them: * Each element has a typical oxidation state, like +1 for hydrogen in most compounds or -2 for oxygen in H2O. * Changes in oxidation states during a reaction show electron movement. For example, if an oxidation state increases, it means the element has lost electrons, which is known as oxidation.
Conversely, if the oxidation state decreases, the element has gained electrons, known as reduction.
In the given exercises: * In reaction (a), chromium in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) changes its oxidation state from +6 to +3, showing that it is being reduced.
On the other hand, tin in \( \mathrm{Sn}^{2+} \) changes from +2 to +4, indicating oxidation.
* For reaction (b), iron in \( \mathrm{FeS} \) is oxidized as its oxidation state goes from 0 to +3.
Whereas nitrogen in \( \mathrm{NO}_{3}^{-} \) is reduced, moving from +5 to +2.
By determining these changes, we can better understand which elements are losing or gaining electrons.

Oxidizing Agent

In a redox reaction, the oxidizing agent plays a crucial role. It accepts electrons and gets reduced in the process. To remember this, think of the oxidizing agent as the electron taker. Here’s a way to identify an oxidizing agent:* Look for the substance that has a decrease in oxidation state. This shows it gains electrons.* The element that undergoes reduction in a reaction is the oxidizing agent.
In reaction (a) of the exercise, \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) serves as an oxidizing agent. It accepts electrons from \( \mathrm{Sn}^{2+} \) and undergoes a reduction in its oxidation state from +6 in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) to +3 in \( \mathrm{Cr}^{3+} \).
Similarly, in reaction (b), \( \mathrm{NO}_{3}^{-} \) is the oxidizing agent. It reduces its oxidation state from +5 to +2, thereby gaining electrons.
Understanding oxidizing agents is key to deciphering the electron flow in chemical reactions.

Reducing Agent

The reducing agent in a redox reaction is the opposite of the oxidizing agent. It's the species that donates electrons and, as a result, itself gets oxidized. Hence, it can be called the electron giver. Ways to identify a reducing agent include:* Observing the rise in oxidation state which indicates it loses electrons.* The element that undergoes oxidation in the reaction is the reducing agent.
In the case of reaction (a), \( \mathrm{Sn}^{2+} \) acts as the reducing agent. It loses electrons as its oxidation state rises from +2 to +4 when forming \( \mathrm{Sn}^{4+} \).
In reaction (b), \( \mathrm{FeS} \) plays the role of a reducing agent. The iron in \( \mathrm{FeS} \) loses electrons while changing from the 0 oxidation state to a +3 in \( \mathrm{Fe}^{3+} \).
Comprehending the function of reducing agents helps explain which elements in the reaction are giving up their electrons.