The standard electrode potential for Daniell cell is $1.1\ V$. Calculate the standard Gibbs energy for the reaction:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Question

Accepted Answer

> [!TLDR]
> The standard Gibbs energy is directly linked to the cell potential via $\Delta G^\circ = -nFE^\circ$. For the Daniell cell, $n=2$, $F=96485\ 	ext{C mol}^{-1}$, and $E^\circ=1.1\ 	ext{V}$, giving $\Delta G^\circ = -212.3\ 	ext{kJ mol}^{-1}$.

The key insight here is that electrochemical cells convert chemical energy into electrical work. The Nernst equation tells us how cell potential varies with concentration, but at standard conditions, the relationship between Gibbs energy and cell potential is beautifully simple: the maximum electrical work the cell can do equals the decrease in Gibbs energy.

For a spontaneous reaction like this one, $\Delta G^\circ$ must be negative, and the positive cell potential confirms that. The magnitude tells us how much useful work we can extract per mole of reaction.

1. **Identify the number of electrons transferred.**  
   In the reaction $Zn(s) + Cu^{2+}(aq) ightarrow Zn^{2+}(aq) + Cu(s)$, zinc goes from oxidation state 0 to +2 (loses 2 electrons), and copper goes from +2 to 0 (gains 2 electrons). So $n = 2$ moles of electrons are transferred per mole of reaction.

2. **Recall the fundamental relation.**  
   The standard Gibbs energy change is related to the standard cell potential by:
   $$\Delta G^\circ = -nFE^\circ$$
   where $F$ is the Faraday constant ($96485\ 	ext{C mol}^{-1}$), $n$ is the number of electrons, and $E^\circ$ is the standard cell potential.

> [!FORMULA]
> $$\Delta G^\circ = -nFE^\circ$$

3. **Plug in the values.**  
   $n = 2$, $F = 96485\ 	ext{C mol}^{-1}$, $E^\circ = 1.1\ 	ext{V}$.
   $$\Delta G^\circ = -(2)(96485\ 	ext{C mol}^{-1})(1.1\ 	ext{V})$$
   $$\Delta G^\circ = -212267\ 	ext{J mol}^{-1}$$

4. **Convert to kilojoules.** …

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.The standard electrode potential for Daniell cell is $1.1\ V$ . Calculate the standard Gibbs energy for the reaction:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

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: Gibbs Energy from Cell Potential (Nernst–Gibbs Relation)

Key Formula

Step-by-Step Solution

1. Forgetting the Sign Convention for $E^\circ_{\text{cell}}$

2. Using the Wrong Number of Electrons ( $n$ )

3. Using the Wrong Value for Faraday’s Constant ( $F$ )

4. Forgetting the Negative Sign in $\Delta G^\circ = -nFE^\circ$

5. Not Converting Units Correctly (J to kJ)

Q.The standard electrode potential for Daniell cell is 1.1 V1.1\ V1.1 V. Calculate the standard Gibbs energy for the reaction: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

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Q.The standard electrode potential for Daniell cell is $1.1\ V$ . Calculate the standard Gibbs energy for the reaction:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Unlock everything free for 14 days