For the disproportionation reaction 2Cu+(aq) Cu(s) + Cu2+(aq) at 298 K, ln K (where K is the equ

Q: Why is $n=1$ in this reaction?

In the disproportionation of $\text{Cu}^+$, one $\text{Cu}^+$ ion loses an electron to become $\text{Cu}^{2+}$ (oxidation), and another $\text{Cu}^+$ ion gains that exact same electron to become solid $\text{Cu}$ (reduction). Thus, the total number of electrons transferred ($n$) is 1.

Q: What does disproportionation mean?

Disproportionation is a specific type of redox reaction where a single substance is simultaneously oxidized and reduced to form two different products.

Q: Why do we subtract the anode potential from the cathode potential?

Standard reduction potentials measure the tendency of a species to gain electrons. The cathode is where reduction actually occurs, so it has a higher potential. Subtracting the anode's reduction potential gives the net driving force (voltage) of the cell.

Q: How did we get the formula relating $E^\circ_{\text{cell}}$ and $\ln K$?

We linked them through the standard Gibbs free energy change ($\Delta G^\circ$). We know $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ and $\Delta G^\circ = -RT \ln K$. Equating these gives $-nFE^\circ_{\text{cell}} = -RT \ln K$, which rearranges to $\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}$.

Q: What if the cell potential was negative?

A negative standard cell potential would mean the standard Gibbs free energy change is positive, indicating that the forward reaction is non-spontaneous under standard conditions, and the equilibrium constant $K$ would be less than 1.

Question

For the disproportionation reaction $2	ext{Cu}^+(aq) ightleftharpoons 	ext{Cu}(s) + 	ext{Cu}^{2+}(aq)$ at 298 K, $\ln K$ (where $K$ is the equilibrium constant) is ......... $	imes 10^{-1}$.
Given : ($E^\circ_{	ext{Cu}^{2+}/	ext{Cu}^+} = 0.16	ext{ V}$, $E^\circ_{	ext{Cu}^+/	ext{Cu}} = 0.52	ext{ V}$ and $\frac{RT}{F} = 0.025$)

Sigma Percentile · Accepted Answer

The problem of finding the equilibrium constant for a disproportionation reaction is a beautiful intersection of thermodynamics and electrochemistry. It requires us to bridge the gap between cell potentials and Gibbs free energy. Let's break down the thought process step-by-step.

### Analyzing the Setup

We are given the disproportionation reaction of copper(I) ions:
$$2	ext{Cu}^+(aq) ightleftharpoons 	ext{Cu}(s) + 	ext{Cu}^{2+}(aq)$$

In this fascinating chemical dance, the $	ext{Cu}^+$ ion is acting as both the oxidizing and the reducing agent. One $	ext{Cu}^+$ ion loses an electron to become $	ext{Cu}^{2+}$ (oxidation), while another $	ext{Cu}^+$ ion gains that exact same electron to become solid $	ext{Cu}$ (reduction).

To find the equilibrium constant ($K$), we first need to determine the standard cell potential ($E^\circ_{	ext{cell}}$). The standard cell potential is the difference between the standard reduction potentials of the cathode and the anode:
$$E^\circ_{	ext{cell}} = E^\circ_{	ext{cathode}} - E^\circ_{	ext{anode}}$$

From the given data, the reduction of $	ext{Cu}^+$ to $	ext{Cu}$ occurs at the cathode ($E^\circ = 0.52	ext{ V}$), and the oxidation of $	ext{Cu}^+$ to $	ext{Cu}^{2+}$ occurs at the anode. The standard reduction potential for the anode half-reaction ($	ext{Cu}^{2+} ightarrow 	ext{Cu}^+$) is given as $0.16	ext{ V}$.

Substituting these values into our equation:
$$E^\circ_{	ext{cell}} = 0.52	ext{ V} - 0.16	ext{ V} = 0.36	ext{ V}$$

### The Master Equation

Now that we have the standard cell potential, how do we connect it to the equilibrium constant? The bridge between these two worlds is the standard Gibbs free energy change ($\Delta G^\circ$).

We know two fundamental thermodynamic relationships:
1. $\Delta G^\circ = -nFE^\circ_{	ext{cell}}$
2. $\Delta G^\circ = -RT \ln K$

By equating these two expressions, we establish a direct relationship between the cell potential and the equilibrium constant:
$$-nFE^\circ_{	ext{cell}} = -RT \ln K$$

Rearranging this to solve for $\ln K$, we get:
$$\ln K = \frac{nFE^\circ_{	ext{cell}}}{RT}$$

### Final Calculation

Let's carefully substitute our known values into this master equation.

Since one electron is transferred in each half-reaction, the number of moles of electrons transferred ($n$) is $1$. We are conveniently given the value of the ratio $\frac{RT}{F}$ as $0.025$.

Plugging everything in:
$$\ln K = \frac{1 	imes 0.36}{0.025}$$

Dividing $0.36$ by $0.025$ yields:
$$\ln K = 14.4$$

The question specifically asks for the answer in the format of a number multiplied by $10^{-1}$. We can easily rewrite $14.4$ to match this format:
$$14.4 = 144 	imes 10^{-1}$$

Thus, our final integer answer is **144**.

Enter Numerical Value:

Visualized Solution

Disproportionation Reaction

Standard Cell Potential $E_{cell}^{\circ}$

Identifying Cathode and Anode

Calculating $E_{cell}^{\circ}$

Relating $E_{cell}^{\circ}$ and $ln K$

Substituting Values

Calculating $ln K$

Final Answer Format

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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