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

Question

Accepted Answer

> [!TLDR]
> The equilibrium constant $K_c$ is found from the Nernst equation at equilibrium: $E^\circ_{cell} = \frac{0.059}{n} \log K_c$. For this reaction, $n=2$ and $E^\circ_{cell}=0.46\ 	ext{V}$, giving $\log K_c = 15.6$ and $K_c = 3.92 	imes 10^{15}$.

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:
$$E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q$$
where $n$ is the number of electrons transferred and $Q$ is the reaction quotient. At equilibrium, $E_{cell}=0$ and $Q=K_c$, so:
$$0 = E^\circ_{cell} - \frac{0.059}{n} \log K_c \quad \Rightarrow \quad \log K_c = \frac{n E^\circ_{cell}}{0.059}$$

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:
   $	ext{Cu}(s) + 2	ext{Ag}^+(aq) ightarrow 	ext{Cu}^{2+}(aq) + 2	ext{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:
   $$\log K_c = \frac{n E^\circ_{cell}}{0.059} = \frac{2 	imes 0.46}{0.059}$$

3. **Calculate $\log K_c$.**
   $$2 	imes 0.46 = 0.92$$
   $$\frac{0.92}{0.059} = 15.59 \approx 15.6$$
   So $\log K_c \approx 15.6$.

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

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

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

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

Method: Nernst Equation at Equilibrium

Steps

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

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

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

4. Forgetting to convert temperature or using the wrong constant

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

6. Calculation errors in the final step …

Q.Calculate the equilibrium constant of the reaction: Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s) Ecell∘=0.46 VE^\circ_{cell} = 0.46\ VEcell∘​=0.46 V

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

Unlock everything free for 14 days