Problem 1 - General Chemistry: Principles and Modern Applications

Question

From the observations listed, estimate the value of \(E^{\circ}\) for the half- reaction \(\mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}(\mathrm{s})\) (a) The metal M reacts with HNO \(_{3}(\text { aq })\), but not with \(\mathrm{HCl}(\mathrm{aq}) ; \mathrm{M}\) displaces \(\mathrm{Ag}^{+}(\mathrm{aq}),\) but not \(\mathrm{Cu}^{2+}(\mathrm{aq})\) (b) The metal \(M\) reacts with \(\mathrm{HCl}(\mathrm{aq}),\) producing \(\mathrm{H}_{2}(\mathrm{g}),\) but displaces neither \(\mathrm{Zn}^{2+}(\text { aq })\) nor \(\mathrm{Fe}^{2+}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The standard electrode potential for the half-reaction of M lies between that of \(H_2/H^+\) and \(Ag^+/Ag\), or \(H_2/H^+\) and \(Zn^{2+}/Zn\) or \(Fe^{2+}/Fe\). Precise ranking needs additional information.

Step 1: Analyzing Part (a)

The metal M reacts with nitric acid (HNO3), but not with hydrochloric acid (HCl). This suggests that the metal M can reduce H+ ions to hydrogen gas. However, not reacting with HCl means it does not have a higher reactivity than hydrogen. Therefore, it must lie below hydrogen in electrochemical series. Additionally, M can displace silver ions (Ag+), but not copper ions (Cu2+). This suggests that it lies between silver and copper. So, it can be deducted that M's \(E^{\circ}\) value lies between that of \(H_2/H^+\) and \(Ag^+/Ag\).

Step 2: Analyzing Part (b)

In this case, M reacts with hydrochloric acid (HCl) to produce hydrogen gas. This suggests a higher reactivity than hydrogen, implying it is above hydrogen in the electrochemical series. However, it can't displace \(Zn^{2+}\) or \(Fe^{2+}\), suggesting its position below zinc and iron in the series. So, M's \(E^{\circ}\) value lies between that of \(H_2/H^+\) and \(Zn^{2+}/Zn\) or \(Fe^{2+}/Fe\).

Step 3: Concluding from Part (a) and (b)

Combining the conclusions from part (a) and part (b), the \(E°\) for the half-reaction of M must lie between that of \(H_2/H^+\) and \(Ag^+/Ag\), or \(H_2/H^+\) and \(Zn^{2+}/Zn\) or \(Fe^{2+}/Fe\). Deciding the exact value or narrower range requires additional data or observations.

Key Concepts

Reactivity of Metals

Reactivity of metals hints at how easily a metal can undergo a chemical reaction. In simple terms, it's about how keen a metal is to join forces with another element.
If you imagine superheroes, more reactive metals are quicker to jump into action! They can easily displace less reactive heroes.
Think of a metal's reactivity as its eagerness to form bonds. More reactive metals often whisk away less reactive elements in a redox reaction. In the exercise, metal M has a particular position in the reactivity series, which helps us predict its behavior in reactions. A super important tool related to this is the reactivity series of metals. It's like a ranking list or a scoreboard that tells us which metals react easier than others. Knowing where a metal sits in this list helps predict what happens during reactions, especially in displacement reactions.

• A more reactive metal will easily react with acids, releasing hydrogen gas.
• It can also push out a less reactive metal from a solution, just like metal M displaces silver ions but not copper ions.
• The position of a metal in this series is crucial for understanding the possible redox reactions it might undergo.

Understanding reactivity is key for making batteries, rust prevention, and even creating fireworks! Reactivity isn't just theory—it's important for real-life applications too.

Standard Electrode Potential

Standard electrode potential, symbolized by \( E^{\circ} \), is like a rating that tells us a lot about a metal's behavior in electrochemistry. It's measured in volts and is a way to gauge a metal's tendency to lose or gain electrons when paired with a standard hydrogen electrode.
Imagine \( E^{\circ} \) as a star rating system for metals—higher positive ratings mean a metal strongly wants to get electrons, while negative ratings suggest it's eager to give them away!
In the exercise, by comparing the \( E^{\circ} \) values for different metal combinations, we can estimate where metal M stands in terms of its standard electrode potential.

• Higher \( E^{\circ} \) values (more positive) indicate non-reactive metals that like to "hold" electrons.
• Lower \( E^{\circ} \) values (more negative) suggest reactive metals more inclined to "give away" electrons.

The electrochemical series is essentially a list arranged by \( E^{\circ} \), helping predict who in a pair will give or receive electrons during a redox reaction. For our exercise:

• Metal M's participation in reactions with hydrochloric acid and its ability to displace some ions indicate where it might sit relative to known metals in the series.
• Aligning metal M's behaviors against known potentials for hydrogen, zinc, and others helps clarify its position.

Exploring \( E^{\circ} \) values isn’t just about experiments; these values guide innovations like creating efficient electrochemical cells and reducing undesired corrosion.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, revolve around the exchange of electrons. They're like a game of catch using electrons!
Imagine one substance passing electrons to another—one is the sender (oxidized), and one is the receiver (reduced).
These reactions are fundamental and power everything from your phone’s battery to biological processes. In the exercise, understanding the redox behavior helps determine how metal M reacts with other substances. This can guide us to find its exact position concerning its standard electrode potential.

• In redox reactions, oxidation involves losing electrons, while reduction means gaining them.
• The substance that loses electrons is the reducing agent, and the one gaining them is the oxidizing agent.
• Identifying who plays which role in our chemical story is key—like figuring out positions in a sports match!

For metal M, its ability to "kick out" certain metal ions in a solution directly relates to these redox principles. Knowing which ions it can displace tells us about its strength as a reducing agent. The beauty of redox reactions lies in their universality. They help us harness energy, develop new compounds, and understand nature’s complex workings. So, cracking the code of redox reactions is a gateway to much more than electrochemical series; it's exploring the very foundation of chemistry!