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Chapter 1: Problem 32

In a voltaic cell with a Cu(s) \(| \mathrm{Cu}^{2+}\) cathode and \(\mathrm{a} \mathrm{Pb}^{2+} | \mathrm{Pb}\) (s) anode, increasing the concentration of \(\mathrm{Pb}^{2+}\) causes the voltage to decrease. What is the reason for this? (A) The value for Q will increase, causing the cell to come closer to equilibrium. (B) The solution at the anode becomes more positively charged, leading to a reduced electron flow. (C) The reaction will shift to the right, causing a decrease in favor ability. (D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.

Short Answer

Expert verified
The correct answer is (A): The value for Q will increase, causing the cell to come closer to equilibrium. This is because increasing the concentration of \(\mathrm{Pb}^{2+}\) ions increases the reaction quotient \(Q\), which according to the Nernst equation, leads to a decrease in cell potential, hence causing a decrease in voltage.

Step by step solution

01

Understand the system

In the given voltaic cell, copper (Cu) functions as the cathode, where reduction occurs, and lead (Pb) functions as the anode, where oxidation occurs. The concentration of \(\mathrm{Pb}^{2+}\) ions in the anode compartment is being increased according to the problem.
02

Consider changes in Q and cell potential

According to Le Chatelier's Principle, the system will respond to reduce the change and restore equilibrium. Here, as we increase the concentration of \(\mathrm{Pb}^{2+}\) ions, the reaction will shift in the direction that consumes \(\mathrm{Pb}^{2+}\) ions. Additionally, as the concentration of \(\mathrm{Pb}^{2+}\) ions increases, the reaction quotient \(Q\) also increases as \(Q\) is the ratio of the concentrations of the products over the reactants in a reaction at a specific moment. When \(Q\) increases, according to the Nernst equation, cell potential reduces.
03

Link the changes to voltage

We know that a higher concentration of ions tends to decrease cell potential, which in turn reduces the electromotive force (or voltage) of the cell. Therefore, in the given voltaic cell, increasing the concentration of \(\mathrm{Pb}^{2+}\) ions decreases the voltage of the cell.
04

Determine the correct answer

After understanding the principles at play and their effects on the voltaic cell, we can definitively say that the correct answer is (A): The value for Q will increase, causing the cell to come closer to equilibrium.

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