Problem 31 - Chemistry The Science in Context

Question

If a piece of silver is placed in a solution in which \(\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) will the following reaction proceed spontaneously? $$2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s)$$

Step-by-Step Solution

Answer

$$2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s)$$ Answer: No, the reaction will not proceed spontaneously under these conditions, as the cell potential (E_cell) is calculated to be -0.46 V, which is less than 0.

Step 1: 1. Write the half-cell reactions

For this calculation, we need to write the half-cell reactions for both silver and copper ions: $$\mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}(s)$$ $$\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}(s)$$

Step 2: 2. Identify the standard reduction potentials

Next, we should look up the standard reduction potentials for both of these half-cell reactions. The standard reduction potentials (E°) are: Ag+: E° = +0.80 V Cu2+: E° = +0.34 V

Step 3: 3. Write the overall reaction equation

Now, we can write the overall reaction equation by combining the half-cell reactions. Note that we need to multiply the silver half-cell reaction by 2: $$2 \mathrm{Ag}^+ + 2e^- \rightarrow 2\mathrm{Ag}(s)$$ $$\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}(s)$$ Adding these together: $$2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s)$$

Step 4: 4. Calculate the cell potential

The Nernst equation enables us to compute the cell potential (E_cell) under non-standard conditions. However, it is essential to note that our given concentrations of Ag+ and Cu2+ are 1.00 M, which are standard conditions. Therefore, we can use the standard reduction potentials directly to calculate the cell potential E_cell: E_cell = E°(reduction) - E°(oxidation) Since Ag is being oxidized and Cu2+ is being reduced, we can plug in our standard potentials: E_cell = (+0.34 V) - (+0.80 V) = -0.46 V

Step 5: 5. Determine the spontaneity of the reaction

If E_cell > 0, then the reaction is considered spontaneous. If E_cell < 0, then the reaction is not spontaneous. In our case: E_cell = -0.46 V Since E_cell < 0, the reaction will not proceed spontaneously under these conditions.

Key Concepts

Understanding Half-Cell Reactions

Half-cell reactions are the fundamental building blocks of electrochemical cells. These are the individual reactions that occur at each electrode in a galvanic or voltaic cell, which efficiently converts chemical energy into electrical energy. Each half-cell reaction involves either a reduction or an oxidation process.

• Reduction occurs when a substance gains electrons.
• Oxidation happens when a substance loses electrons.

In our example, we have two half-cell reactions:

• The reduction of silver: \[ \mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag}(s) \]
• The reduction of copper (which is actually oxidizing in context): \[ \mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}(s) \]

These half-reactions show us how electrons move in the system, which is crucial for determining the feasibility of the entire electrochemical reaction. By understanding half-cell reactions, students can easily break down complex overall reactions into simpler, more manageable parts.

The Role of Standard Reduction Potentials

Standard reduction potentials (\( E° \)) are essential in understanding how likely a half-cell reaction is to occur under standard conditions. These values, typically measured in volts, provide a relative measurement of a substance's ability to be reduced. Elements or compounds with higher \( E° \) values are more likely to gain electrons (and thus be reduced) compared to those with lower values.

In our exercise, the standard reduction potentials help indicate:

• Ag+ has an \( E° \) of +0.80 V, suggesting it has a strong tendency to be reduced to Ag(s).
• Cu2+ has an \( E° \) of +0.34 V, showing it's less likely to be reduced under the same conditions.

Understanding these values allows us to predict how electrons will flow in a galvanic cell. By setting up an accurate portrayal of the half-reactions and their respective potentials, students can predict the direction of the electron flow, and thus, assess the spontaneity of the overall electrochemical reaction.

Understanding Cell Potential

The cell potential, also known as the electromotive force (EMF) or \( E_{cell} \), is a measure of the voltage difference between two half-cells in an electrochemical cell. It essentially determines whether a reaction will proceed spontaneously. This potential is calculated using the standard reduction potentials of the half-cells involved. The formula is:
\[E_{cell} = E°_{reduction} - E°_{oxidation}\]
In our example:

• The copper side has an \( E°_{reduction} \) of +0.34 V because it gains electrons.
• The silver side, when reversed, shows an \( E°_{oxidation} \) of +0.80 V.

The overall cell potential is calculated as:\[E_{cell} = 0.34 \, V - 0.80 \, V = -0.46 \, V\]
A negative cell potential indicates that the reaction is non-spontaneous under standard conditions. Positive cell potential values mean the reaction could spontaneously occur. Understanding the cell potential is crucial for predicting the feasibility of electrochemical reactions and determining whether external energy would be required to drive a non-spontaneous process.