Problem 54 - General Chemistry: Principles and Modern Applications

Question

The following concentration cell is constructed. \(\mathrm{Ag} | \mathrm{Ag}^{+}\left(0.10 \mathrm{M}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, 0.10 \mathrm{M} \mathrm{CN}^{-}\right)\) $$\| \mathrm{Ag}^{+}(0.10 \mathrm{M}) | \mathrm{Ag}$$ If \(K_{f}\) for \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) is \(5.6 \times 10^{18},\) what value would you expect for \(E_{\text {cell }}\) ? [Hint: Recall that the anode is on the left.]

Step-by-Step Solution

Answer

The expected value for \(E_{\text {cell }}\) is -0.0296 V

Step 1: Identify the half-reactions

Let's identify the half-reactions. For the left half-cell, \(\mathrm{Ag}(\mathrm{CN})_{2}\) is converted into Ag and this half-reaction can be represented as: \[\mathrm{Ag}(\mathrm{CN})_{2}^{-} + e^{-} \rightarrow \mathrm{Ag} + 2 \mathrm{CN}^{-}\] Similarly for the right half cell, we have: \[\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}\]

Step 2: Calculate the half-cell reduction potentials

Let’s understand that the left half-cell is the anode where oxidation occurs and the right half-cell is the cathode where reduction occurs. According to the Nernst equation, the electromotive force (EMF) of a half-cell reaction at non-standard conditions can be represented as: \[E = E^{\circ} - \frac{0.0592}{n} \log Q\] where \(E^{\circ}\) is the standard reduction potential, n is the number of electron transfer, and Q is the reaction quotient. For the left half-cell, we write the reaction quotient (Q) as \([Ag][CN^-]^2/ [Ag(CN)_2^-] = 0.0010\), given that the concentration of Ag, which is a solid, is counted as 1 in the equilibrium expression. For the right half-cell, as the reaction is under standard conditions, \(Q = 1\) and hence the Ecell is simply the standard reduction potential. Note that the value of \(E^{\circ}\) for both half reactions is given as zero since Ag+ ions are involved in both half-cells.

Step 3: Plug the values into the Nernst equation

Now, we can plug values into the Nernst equation to get E for both half-cells. For left half cell, we have \(E_{\text{left}} = 0 - \frac{0.0592}{1} \log (0.0010) = 0.0296 V\). For the right half cell under standard conditions, we have \(E_{\text{right}} = 0 V\). The overall cell potential can be calculated as the difference between the reduction potentials at the cathode and anode, i.e. \(E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = 0 - 0.0296 = -0.0296 V\).

Step 4: Convert the final answer to required unit

The final electromotive potential should be determined in Volt (V). Note that the negative sign in the answer indicates the cell is not spontaneous in the given direction. In order to get a positive cell potential, the cell would need to be reversed.

Key Concepts

Nernst equation

The Nernst equation is a fundamental formula used to determine the electromotive force (EMF) of a galvanic cell under non-standard conditions. When dealing with concentration cells, like the problem described, the Nernst equation becomes essential to calculate the cell potential. It takes into account the concentrations of ions participating in the electrochemical reaction.
In mathematical terms, the Nernst equation is written as:

\[E = E^{\circ} - \frac{0.0592}{n} \log Q\]

Here's a quick breakdown of the terms:

• \(E\) is the EMF at non-standard conditions.
• \(E^{\circ}\) is the standard EMF, which is zero for a concentration cell.
• \(n\) represents the number of moles of electrons transferred in the reaction.
• \(Q\) is the reaction quotient, which reflects the ratio of concentrations of products to reactants.

In the context of the exercise, the Nernst equation helped determine the EMF value for the half-cells with different concentrations, ultimately leading to a calculation for the overall cell potential.

half-cell reactions

Half-cell reactions are the foundation of understanding how a concentration cell functions. In an electrochemical cell, two half-cells work together, each involving a different reaction. These are categorized as:

• The **oxidation reaction**, which occurs at the anode and involves the loss of electrons.
• The **reduction reaction**, taking place at the cathode, where electrons are gained.

For the concentration cell in the exercise, two half-cell reactions were considered. In the left half-cell, the species \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) is oxidized to \(\mathrm{Ag}\), giving off electrons, as described by the equation:

\[\mathrm{Ag}(\mathrm{CN})_{2}^{-} + e^{-} \rightarrow \mathrm{Ag} + 2 \mathrm{CN}^{-}\]

Meanwhile, on the right-hand side, the reduction reaction involves \(\mathrm{Ag}^{+}\) ions gaining electrons to form \(\mathrm{Ag}\):

\[\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}\]

By separating these reactions into half-cell equations, it is easier to analyze and calculate the contributions to the cell's overall electromotive force.

electromotive force

Electromotive force (EMF) is essentially the voltage produced by a cell when no current flows. It indicates the cell's ability to drive electrons through the circuit. For concentration cells, the EMF depends on the difference in concentrations of ions in each half-cell.
Notice from the problem that the concentration of \(\mathrm{Ag}^+\) ions was the same in each half-cell, influencing the calculated EMF values. For the left half-cell, we factored in a \(Q\) value distinct from standard conditions, leading to an EMF that isn't zero.

This difference between EMF values is translated into cell potential using the formula:

\[E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\]

In this exercise, this calculated EMF was negative, indicating a non-spontaneous configuration in the set direction. Electromotive force provides insight into how changes in ionic concentrations alter the energy output of a cell, a crucial point when exploring varied electrochemical systems.