Numerical/Formula

1. TITLE

E°cell vs. Equilibrium Constant (Kc) numerical — The method that works every time

2. PROBLEM TYPE INTRO

This numerical falls under the category of Electrochemistry equilibrium problems. It requires calculating the equilibrium constant ($K_c$) from the standard cell potential ($E^\circ_{\text{cell}}$) at 298 K. We use the Nernst equation applied at equilibrium. The typical mistake students make is misidentifying the number of electrons transferred ($n$) or stumbling when taking the antilog without log tables.

3. MCQ

Question: Calculate the equilibrium constant ($K_c$) for the reaction: $\text{Cu}(s) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Ag}(s)$ at 298 K, given that the standard cell potential $E^\circ_{\text{cell}}$ is $0.46\text{ V}$.

Options:
(A) $3.92 \times 10^{15}$
(B) $3.92 \times 10^{7}$
(C) $3.92 \times 10^{-15}$
(D) $1.56 \times 10^{1}$

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* Reaction: $\text{Cu}(s) + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Ag}(s)$
* Standard cell potential, $E^\circ_{\text{cell}} = 0.46\text{ V}$
* Temperature, $T = 298\text{ K}$

To Find:
* Equilibrium constant ($K_c$)

Formula:
* $E^\circ_{\text{cell}} = \frac{0.059\text{ V}}{n} \log K_c$
* Explain when to use this formula: Use this specific simplified formula when you need to find the relationship between the standard cell potential and the equilibrium constant at exactly 298 K.

Step 1: Identify the number of moles of electrons transferred ($n$)
Split the reaction into half-reactions:
Oxidation: $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2e^-$
Reduction: $2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag}(s)$
The number of electrons lost and gained is 2, so $n = 2$.

Step 2: Substitute the values into the formula and solve for $\log K_c$

$$0.46\text{ V} = \frac{0.059\text{ V}}{2} \log K_c$$

 

$$\log K_c = \frac{0.46 \times 2}{0.059}$$

 

$$\log K_c = \frac{0.92}{0.059} \approx 15.593$$


We can round this to $15.6$.

 

Step 3: Solve for $K_c$ by taking the antilog (without log tables)

$$K_c = \text{antilog}(15.6) = 10^{15.6}$$


Split the exponent into an integer and a decimal:

$$10^{15.6} = 10^{15} \times 10^{0.6}$$


Recall basic log values: $\log 2 \approx 0.30$, $\log 3 \approx 0.48$, $\log 4 \approx 0.60$.
Since $\log 4 \approx 0.60$, it means $10^{0.60} \approx 4$.
Therefore, $10^{0.6}$ is just under 4, which perfectly matches $3.92$.

$$K_c = 3.92 \times 10^{15}$$

 

Answer: $3.92 \times 10^{15}$ (Option A)

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Misidentifying 'n' (number of electrons transferred). Students often see the $+1$ charge on $\text{Ag}^+$ and mistakenly use $n = 1$, yielding $\log K_c = 7.8$, leading to Option B ($3.92 \times 10^7$).
  • ❌ Mistake 2: Using the wrong sign in the potential. Students might mistakenly add a negative sign thinking of the Gibbs energy formula, which would yield the impossible Option C ($3.92 \times 10^{-15}$).
  • ❌ Mistake 3: Stopping at the log value. Forgetting to take the antilog entirely and just selecting $15.6$ as the final answer (Option D).

6. QUICK FORMULA REFERENCE

Formulas Used:
* $E^\circ_{\text{cell}} = \frac{0.059\text{ V}}{n} \log K_c$ — Use when calculating $K_c$ from standard cell potential at 298 K.
* $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059\text{ V}}{n} \log Q$ — Use when the cell is not at equilibrium and you need to calculate the non-standard cell potential.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: Calculating the equilibrium constant for a Daniell Cell where $E^\circ = 1.1\text{ V}$ and $n = 2$."

8. NCERT REFERENCE

Based on: NCERT Class 12, Chapter 2, Example 2.2.

9. CTA

More solved numericals → Redox Reactions & Electrochemistry Complete Guide
Practice problems → Mock exams