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Chapter 10: Problem 24

Consider a battery made from lead, \(\mathrm{Pb}\), and copper, \(\mathrm{Cu}\), along with \(\mathrm{Pb}^{2+}\) ions and \(\mathrm{Cu}^{2+}\) ions. In such a battery, what is the oxidizing agent, and what is the reducing agent? (Hint: Start by consulting the EMF series, and determine which metal oxidizes and which metal ion gets reduced.)

Short Answer

Expert verified
In the battery made from lead (Pb) and copper (Cu) along with \(\mathrm{Pb}^{2+}\) ions and \(\mathrm{Cu}^{2+}\) ions, the oxidizing agent is \(\mathrm{Cu^{2+}}\) ions, and the reducing agent is lead (Pb). This is because lead is higher in the EMF series, so it gets oxidized (loses electrons) and forms \(\mathrm{Pb^{2+}}\) ions, while copper ions gain electrons (get reduced) and become copper (Cu) atoms.

Step by step solution

01

Consult the EMF Series

Refer to the Electromotive Force (EMF) series to observe the reactivity of metals and their ions. The position of a metal in the EMF series determines its strength as a reducing agent.
02

Find the positions of lead and copper in the EMF series

Look for lead (Pb) and copper (Cu) in the EMF series. Remember that elements in the EMF series are arranged according to their reducing power, with the most reactive metals at the top (that is, those which are most likely to lose electrons or be oxidized) and the least reactive at the bottom (those less likely to lose electrons or be oxidized). In the EMF series, you will find that lead (Pb) appears higher than copper (Cu).
03

Determine which metal gets oxidized and which ion gets reduced

Based on the positions of lead and copper in the EMF series, we can conclude that lead (Pb) is more likely to lose electrons (get oxidized) than copper (Cu). Therefore, lead (Pb) will get oxidized (lose electrons) and form lead ions, \(\mathrm{Pb^{2+}}\). On the other hand, copper ions, \(\mathrm{Cu^{2+}}\), will gain electrons (get reduced) and become copper (Cu) atoms.
04

Identify the oxidizing agent and the reducing agent

Now that we know which metal gets oxidized and which ion gets reduced, we can identify the oxidizing and reducing agents. The oxidizing agent is the substance that causes the oxidation (that is, it gains electrons). In this case, the oxidizing agent is \(\mathrm{Cu^{2+}}\) ions because they accept the electrons from lead (Pb). The reducing agent is the substance that causes the reduction (that is, it loses electrons). In this case, the reducing agent is lead (Pb) because it loses electrons to form \(\mathrm{Pb^{2+}}\) ions. To summarize, in the given battery: - The oxidizing agent is \(\mathrm{Cu^{2+}}\) ions. - The reducing agent is lead (Pb).

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