Problem 27 - Chemistry: An Atoms-Focused Approach

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 procecd spontancously? $$ 2 \Lambda g(s)+\mathrm{Cu}^{2+}(a q) \rightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) $$

Step-by-Step Solution

Answer

Reaction: $$ 2 \mathrm{Ag}(s) + \mathrm{Cu}^{2+}(aq) \rightarrow 2 \mathrm{Ag}^{+}(aq) + \mathrm{Cu}(s) $$ Given conditions: \([\mathrm{Ag}^+]=1\,\text{M}\), \([\mathrm{Cu}^{2+}]=1\,\text{M}\) Answer: No, the reaction will not occur spontaneously under the given conditions.

Step 1: Identify the half-reactions and the standard reduction potentials

The given reaction is: $$ 2 \mathrm{Ag}(s) + \mathrm{Cu}^{2+}(aq) \rightarrow 2 \mathrm{Ag}^{+}(aq) + \mathrm{Cu}(s) $$ The half-reactions are: 1. Oxidation of silver: \(\mathrm{Ag}(s) \rightarrow \mathrm{Ag}^{+}(aq) + e^-\) 2. Reduction of copper: \(\mathrm{Cu}^{2+} (aq) + 2e^- \rightarrow \mathrm{Cu}(s)\) Consulting a standard reduction potential table, we find that: 1. For silver: \(E^0_{\mathrm{Ag}^{+}/\mathrm{Ag}} = +0.80\,\mathrm{V}\) 2. For copper: \(E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34\,\mathrm{V}\)

Step 2: Calculate the cell potential under standard conditions

To calculate the cell potential under standard conditions, we can use the following formula: $$ E^0_\text{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}} $$ In this case, the cathode is the copper half-cell, and the anode is the silver half-cell, so: $$ E^0_\text{cell} = (+0.34\,\mathrm{V}) - (+0.80\,\mathrm{V}) = -0.46\,\mathrm{V} $$

Step 3: Calculate the cell potential under the given conditions using the Nernst equation

The Nernst equation allows us to calculate the cell potential under non-standard conditions. The equation is: $$ E_\text{cell} = E^0_\text{cell} - \frac{RT}{nF} \ln Q $$ Where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(F\) is the Faraday constant, \(n\) is the number of moles of electrons transferred, and \(Q\) is the reaction quotient. Assuming the temperature is 298 K, the Nernst equation becomes: $$ E_\text{cell} = -0.46\,\text{V} - \frac{8.314\,\text{J/mol}\cdot\text{K} \cdot 298\,\text{K}}{2 \cdot 96485\,\text{C/mol}} \ln Q $$ We are given the initial concentrations of the ions: \([\mathrm{Ag}^+]=1\,\text{M}\) and \([\mathrm{Cu}^{2+}]=1\,\text{M}\). The reaction quotient, \(Q\), for the reaction is: $$ Q = \frac{[\mathrm{Ag}^+]^2}{[\mathrm{Cu}^{2+}]} $$ Plugging in the given concentrations, we have: $$ Q = \frac{(1\,\text{M})^2}{(1\,\text{M})} = 1 $$ Now we can plug in the value of \(Q\) into the Nernst equation: $$ E_\text{cell} = -0.46\,\text{V} - \frac{8.314\,\text{J/mol}\cdot\text{K} \cdot 298\,\text{K}}{2 \cdot 96485\,\text{C/mol}} \ln (1) = -0.46\,\mathrm{V} $$

Step 4: Determine if the reaction is spontaneous

The cell potential under the given conditions is -0.46 V. Since the cell potential is negative, the reaction is not spontaneous. Therefore, the given reaction will not proceed spontaneously under the given conditions.

Key Concepts

Standard Reduction Potential

In electrochemistry, the Standard Reduction Potential is a crucial concept that helps determine how likely a substance is to gain electrons and be reduced. It is denoted by \( E^0 \) and measured in volts (V). These values are determined under standard conditions: concentrations of 1 M, pressure of 1 atm, and a temperature of 298 K.

Each half-reaction has its associated standard reduction potential, which reflects its tendency to undergo reduction. For example, in the given reaction, silver \((\text{Ag}^+)\) has a standard reduction potential \( E^0_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ V} \), while copper \((\text{Cu}^{2+})\) has \( E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \).

• Higher \( E^0 \) values mean a greater tendency to be reduced.
• The copper ion reduction occurs at the cathode (where reduction takes place).
• The silver undergoes oxidation at the anode.

The potential difference between the cathode and anode reflects the overall cell potential under standard conditions and plays a key role in understanding the spontaneity of reactions.

Nernst Equation

The Nernst Equation is a useful tool in electrochemistry to calculate the cell potential under non-standard conditions. It adjusts the standard cell potential \( E^0_\text{cell} \) for different concentrations and temperatures, accounting for real-world scenarios. The equation is:
\[E_\text{cell} = E^0_\text{cell} - \frac{RT}{nF} \ln Q\]
Here, \( R \) represents the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient, which is the ratio of the products' concentrations to the reactants' concentrations raised to the power of their coefficients.

In our scenario, the initial concentrations of silver and copper ions are both 1 M, making \( Q = 1 \). Because \( \ln(1) = 0 \), the cell potential remains \( E_\text{cell} = -0.46 \text{ V} \), as calculated earlier. This illustrates that even with balanced concentrations, the reaction's drives may vary if concentration values change significantly.

Spontaneity of Reaction

Understanding whether a chemical reaction will proceed on its own without external influence is a key concept in chemistry. The spontaneity of a reaction can be determined by the sign of the cell potential \( E_\text{cell} \). If \( E_\text{cell} > 0 \), the reaction is spontaneous; this means the products have a lower energy state than the reactants, and the reaction can proceed on its own. Conversely, \( E_\text{cell} < 0 \) indicates that the reaction is non-spontaneous.

In our example, the calculated \( E_\text{cell} \) is \(-0.46 \text{ V} \). A negative value indicates that the reaction is not spontaneous under these conditions. This can be influenced by altering concentrations, temperature, or using an external source of energy, such as an electric current, to drive the reaction.

• Spontaneity means the process can occur without energy input.
• Non-spontaneous reactions may need additional energy or altered conditions.
• It's possible to drive some non-spontaneous reactions using electrolysis.

Understanding these principles helps in predicting and manipulating chemical reactions efficiently.