Question

The position of some metals in the electrochemical series in decreasing electropositive character is given as \(\mathrm{Mg}>\mathrm{Al}>\mathrm{Zn}>\mathrm{Cu}>\mathrm{Ag}\). What will happen if \(\mathrm{a}\) copper spoon is used to stir a solution of aluminium nitrate? (1) The spoon will get coated with aluminium (2) The alloy of copper and aluminium is formed(3) The solution becomes blue (4) There is no reaction

Short answer

There is no reaction.

Step-by-step solution

Understand the Electrochemical Series

The electrochemical series lists metals in order of their electropositivity, from more reactive to less reactive. Magnesium (Mg) is the most electropositive among these metals, and silver (Ag) is the least electropositive.

Identify the Relevant Metals

For this problem, we are concerned with copper (Cu) and aluminium (Al). According to the series, aluminium is more electropositive than copper, meaning aluminium has a greater tendency to lose electrons and form positive ions.

Determine Possible Reactions

When a metal is placed in a solution of another metal's ions, a reaction occurs if the solid metal is more electropositive than the metal ion in solution. Copper is less electropositive than aluminium, meaning copper ions will not displace aluminium ions from the solution.

Analyze the Situation

Since copper is less electropositive than aluminium, copper will not gain electrons from aluminium ions in the solution. Therefore, aluminium will not coat the copper spoon, no alloy will form, and the solution’s color will not change due to these properties.

Conclusion

Given that copper is less electropositive than aluminium and cannot displace aluminium from aluminium nitrate, there will be no reaction when a copper spoon is used to stir a solution of aluminium nitrate.

Key concepts

Electropositivity

Electropositivity refers to a metal’s ability to lose electrons and form positive ions. Higher electropositivity means a metal easily loses electrons, making it more reactive. For example:
- Magnesium (Mg) is highly electropositive.
- Silver (Ag) is less electropositive.
In the electrochemical series, metals are listed in decreasing order of electropositivity. A metal listed higher can displace ions of metals listed lower in the series from solutions. This is why we see metals like Mg react more readily compared to metals like Cu.
In the given problem:
- Aluminium (Al) is more electropositive than Copper (Cu).
Thus, copper cannot displace aluminium when a copper spoon stirs aluminium nitrate solution. This understanding helps in analyzing displacement reactions.

Displacement Reactions

Displacement reactions occur when a more electropositive metal displaces a less electropositive metal from its solution. Consider the electrochemical series:
\(\text{Mg} > \text{Al} > \text{Zn} > \text{Cu} > \text{Ag}\)
- Here, Mg can displace Al, Zn, Cu, and Ag.
- Al can displace Zn, Cu, and Ag, but not Mg.
In our problem, Copper (Cu) is less electropositive than Aluminium (Al). This means:
- \(\ce{Cu}\) cannot displace \(\ce{Al^{3+}}\) from Aluminium nitrate.
Therefore, no displacement reaction occurs when a copper spoon stirs Al(NO3)3 solution. Understanding displacement reactions helps predict if and how elements will react in solutions based on their positions in the electrochemical series.

Metal Reactivity

The reactivity of metals is closely tied to their electropositivity. More electropositive metals are more reactive. Key points to note about metal reactivity:
- Highly reactive metals like Mg are vigorous in reactions.
- Less reactive metals like Ag have fewer reactions.
In a given series:
- Reactivity decreases from Mg to Ag.
In the context of our problem:
- Since Al is more reactive than Cu, Al ions won't be displaced by Cu.
This principle is consistent in predicting outcomes where the relative positions of metals in the electrochemical series determine the occurrence of reactions. Hence, the copper spoon stirred in aluminium nitrate leads to no reaction due to the lower reactivity of Cu compared to Al.