Q.Calculate the equilibrium constant of the reaction:
$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$
$E^\circ_{cell} = 0.46\ V$

The key idea is that at equilibrium the cell potential is zero, so the Nernst equation reduces to a direct relation between $E^\circ_{cell}$ and the equilibrium constant $K_c$.

Step 1: Write the Nernst equation at equilibrium ($E_{cell} = 0$, $Q = K_c$):

Step 2: Identify $n$, the number of electrons transferred. The reaction $Cu(s) + 2Ag^+ \rightarrow Cu^{2+} + 2Ag(s)$ shows $Cu \to Cu^{2+}$ (loss of 2 electrons) and $2Ag^+ \to 2Ag$ (gain of 2 electrons), so $n = 2$.

Step 3: Rearrange and substitute $E^\circ_{cell} = 0.46\ \text{V}$ and $n = 2$:

Step 4: Take the antilog: …

The key idea is that the Nernst equation connects cell potential to the reaction quotient, and at equilibrium the cell potential becomes zero while the reaction quotient becomes the equilibrium constant. This gives a direct route from $E^\circ_{cell}$ to $K_c$ without any extra data.

The Nernst equation for a cell reaction at 298 K is:

where $n$ is the number of electrons transferred and $Q$ is the reaction quotient. At equilibrium, $E_{cell}=0$ and $Q=K_c$, so:

This is a standard result for electrochemistry problems — it lets you calculate $K_c$ from the standard cell potential alone.

1. Identify $n$, the number of electrons transferred.

   The reaction is:

   $\text{Cu}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + 2\text{Ag}(s)$

   Copper goes from 0 to +2, losing 2 electrons. Each silver ion goes from +1 to 0, gaining 1 electron, and there are two silver ions, so total electrons gained = 2. Hence $n=2$.

2. Write the equilibrium relation.

   Using the formula above:

3. Calculate $\log K_c$.

So $\log K_c \approx 15.6$.

4. Find $K_c$ from $\log K_c$. Since $\log$ here is base 10, $K_c = 10^{\log K_c} = 10^{15.59}$. …

Method: Nernst Equation at Equilibrium

This method uses the relationship between the standard cell potential and the equilibrium constant via the Nernst equation.

Steps

1. Recall the Nernst equation at equilibrium At equilibrium, $E_{cell} = 0$ and $Q = K_c$. The Nernst equation (at 298 K) becomes:

Substituting equilibrium conditions:

2. Rearrange to solve for $\log K_c$

3. Identify $n$ (number of electrons transferred)

   From the reaction:

   • $Cu \rightarrow Cu^{2+} + 2e^-$ (oxidation, loss of 2 electrons)
   • $2Ag^+ + 2e^- \rightarrow 2Ag$ (reduction, gain of 2 electrons)

   So, $n = 2$.

4. Plug in the values

5. Calculate

6. Find $K_c$ …

Here are the common mistakes students make when solving for the equilibrium constant using the Nernst equation for this reaction, along with how to avoid each.

1. Forgetting that $n$ (number of electrons) is not the same as the stoichiometric coefficient

Mistake:

Students often see "2 Ag⁺" and think $n = 2$ — but they forget to check the half-reactions.

How to avoid:

Write the two half-reactions explicitly:

• Oxidation: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$
• Reduction: $2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)$

The electrons cancel: $n = 2$.

Always confirm $n$ by balancing electrons, not by looking at coefficients.

2. Using the wrong form of the Nernst equation at equilibrium

Mistake:

Plugging $E_{cell} = 0$ into the full Nernst equation but forgetting that at equilibrium $Q = K_c$.

How to avoid:

At equilibrium:

• $E_{cell} = 0$
• $Q = K_c$

So the equation becomes:

Rearrange to:

Memorise this shortcut form for equilibrium constant problems.

3. Using the wrong value of $E^\circ_{cell}$ sign

Mistake:

Taking $E^\circ_{cell}$ as negative or mixing up cathode and anode.

How to avoid:

For a spontaneous reaction (which this is, since $E^\circ_{cell} > 0$), $E^\circ_{cell}$ is positive.

Here $E^\circ_{cell} = +0.46\ V$ is given — use it directly.

Always check: if $K_c > 1$, then $E^\circ_{cell} > 0$.

4. Forgetting to convert temperature or using the wrong constant

Mistake:

Using $0.059$ when the temperature is not 298 K, or using $RT/F$ without simplification.

How to avoid:

The factor $0.059$ (equivalently $0.0591$) is valid only at 298 K.

If the temperature changes, use:

For standard problems, assume 298 K unless stated otherwise.

5. Misinterpreting $\log$ vs $\ln$

Mistake:

Using $\ln$ in the simplified equation that uses $\log_{10}$.

How to avoid:

The simplified form uses base-10 log:

If you use $\ln$, the constant changes to $\frac{RT}{F} \approx 0.0257\ V$.

Stick to $\log_{10}$ with $0.059$ for a quick calculation.

6. Calculation errors in the final step …