If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the

Q: Why do we use $\ln K$ instead of $\log_{10} K$?

Because the fundamental thermodynamic formula $\Delta G^\circ = -RT \ln K$ naturally uses the natural logarithm (base $e$). If we used base 10, we would need to introduce the $2.303$ conversion factor.

Q: How did we know $n=2$ for this reaction?

By looking at the oxidation states in the balanced equation: Zinc goes from $0$ to $+2$ (losing 2 electrons), and Copper goes from $+2$ to $0$ (gaining 2 electrons). Thus, $2$ moles of electrons are transferred per mole of reaction.

Q: What does such a large $K$ value ($e^{160}$) actually mean physically?

It means the reaction goes almost entirely to completion. At equilibrium, the concentration of products is astronomically higher than the concentration of reactants.

Q: Why is the cell potential positive?

A positive $E^\circ$ indicates that the redox reaction is spontaneous under standard conditions, which aligns with the large positive equilibrium constant.

Q: Can we use the standard value $R = 8.314$ instead of $8$?

No! The problem explicitly gives $R = 8 \text{ JK}^{-1}\text{mol}^{-1}$ to simplify your calculations. Always use the values provided in the question to arrive at the exact options given.

Question

If the standard electrode potential for a cell is $2	ext{ V}$ at $300	ext{ K}$, the equilibrium constant ($K$) for the reaction,
$	ext{Zn}(s) + 	ext{Cu}^{2+}(aq) ightleftharpoons 	ext{Zn}^{2+}(aq) + 	ext{Cu}(s)$
at $300	ext{ K}$ is approximately
($R = 8	ext{ JK}^{-1}	ext{mol}^{-1}$, $F = 96000	ext{ C mol}^{-1}$)

Options:

(A) $e^{-160}$

(B) $e^{160}$

(C) $e^{-80}$

(D) $e^{320}$

Sigma Percentile · Accepted Answer

### The Bridge Between Voltage and Equilibrium

Imagine a standard Zinc-Copper galvanic cell. The zinc electrode is dipping in zinc ions, and the copper electrode is in copper ions. The voltmeter reads a standard electrode potential of $2	ext{ V}$. But what happens when the battery eventually dies? It reaches equilibrium! To find the equilibrium constant ($K$), we need the bridge that connects thermodynamics to electrochemistry.

### The Master Equation

The standard Gibbs free energy ($\Delta G^\circ$) is the ultimate link. It connects the standard cell potential ($E^\circ$) and the equilibrium constant ($K$) through two fundamental equations:

$$ \Delta G^\circ = -nFE^\circ $$
$$ \Delta G^\circ = -RT \ln K $$

By equating these two expressions, we get our master formula:

$$ -nFE^\circ = -RT \ln K $$
$$ \ln K = \frac{nFE^\circ}{RT} $$

### Gathering the Variables

Now, let's gather our given values. The reaction involves the transfer of $2$ electrons (Zinc goes from $0$ to $+2$, and Copper goes from $+2$ to $0$), so $n = 2$. The standard potential $E^\circ$ is $2	ext{ V}$, temperature $T$ is $300	ext{ K}$, and we have the specific values for the gas constant $R = 8 	ext{ JK}^{-1}	ext{mol}^{-1}$ and Faraday's constant $F = 96000 	ext{ C mol}^{-1}$.

### The Final Calculation

Let's substitute these values into our rearranged formula for the natural log of $K$:

$$ \ln K = \frac{2 	imes 96000 	imes 2}{8 	imes 300} $$

Don't make a silly mistake here with the calculation. The numerator becomes $4 	imes 96000$, which is $384000$. The denominator is $2400$.

$$ \ln K = \frac{384000}{2400} $$

Dividing $384000$ by $2400$, the zeros cancel out nicely, and $3840$ divided by $24$ gives us exactly $160$. So, the natural log of $K$ is $160$.

$$ \ln K = 160 $$

Finally, to isolate $K$, we take the exponential of both sides. This gives us:

$$ K = e^{160} $$

**A massive number!** This indicates the reaction strongly favors the products. Always remember, a positive standard cell potential corresponds to a highly positive equilibrium constant, meaning the forward reaction is highly spontaneous.

Select Answer:

Visualized Solution

\text{The Galvanic Cell}

\text{Relation between } E^\circ \text{ and } K

\text{Identifying Variables}

\text{Substituting Values}

\text{Simplifying the Expression}

\text{Calculating } \ln K

\text{Finding } K

\text{Conclusion}

The Sigma Insight: Electrochemical Cells

The Bridge Between Voltage and Equilibrium

The Master Equation

Gathering the Variables

The Final Calculation

Similar Questions

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