Consider an electrochemical cell: A(s) | A^{n+} (aq, 2M) || B^{2n+} (aq, 1M) | B(s). The value of \D

Q: Why is $\Delta G = 0$ when EMF is zero?

When the cell potential (EMF) is zero, the cell can no longer do electrical work. This means the reaction has reached equilibrium, and the free energy of the system is at a minimum, hence $\Delta G = 0$.

Q: Why did we multiply the anode reaction by 2?

To balance the overall cell reaction, the number of electrons lost at the anode must equal the number gained at the cathode. Since B needs $2n$ electrons and A only gives $n$, we must multiply A's reaction by 2.

Q: Why is the concentration of solids not included in Q?

The activity of pure solids is taken as unity (1) because their concentration (density divided by molar mass) remains constant during the reaction.

Q: Can $\Delta S^\circ$ be negative for a spontaneous reaction?

Yes! The spontaneity of a reaction is determined by $\Delta G$, not just $\Delta S$. A reaction can be spontaneous ($\Delta G < 0$) even if $\Delta S < 0$, provided the reaction is sufficiently exothermic ($\Delta H < 0$) at that temperature.

Q: What does a negative $\Delta S^\circ$ physically mean here?

It means the products are in a more ordered state than the reactants. Even though we produce more aqueous ions, their strong interaction with water molecules (hydration) restricts the movement of the solvent, leading to an overall decrease in randomness.

Question

Consider an electrochemical cell: $A(s) | A^{n+} (aq, 2	ext{M}) || B^{2n+} (aq, 1	ext{M}) | B(s)$. The value of $\Delta H^	heta$ for the cell reaction is twice that of $\Delta G^	heta$ at $300	ext{ K}$. If the emf of the cell is zero, the $\Delta S^	heta$ (in $	ext{J K}^{-1} 	ext{mol}^{-1}$) of the cell reaction per mole of $B$ formed at $300	ext{ K}$ is______.
(Given : $\ln (2) = 0.7$, $R$ (universal gas constant) $= 8.3	ext{ J K}^{-1} 	ext{mol}^{-1}$. $H$, $S$ and $G$ are enthalpy, entropy and Gibbs energy, respectively.)

Sigma Percentile · Accepted Answer

### Decoding the Electrochemical Cell

Imagine you are standing in front of a beautifully constructed electrochemical cell.

On your left, you have electrode $A$ resting in a concentrated $2	ext{M}$ solution of its own ions. On your right, electrode $B$ sits in a $1	ext{M}$ solution.

But here is the twist—the voltmeter connecting them reads exactly zero!

**What does a zero EMF physically mean?** It means the cell has exhausted its ability to do electrical work. The chemical push and pull between the two compartments have perfectly balanced out. The system has reached a state of dynamic equilibrium.

### The Chemical Choreography

Before we dive into the thermodynamics, we must understand the chemical dance happening at the electrodes.

At the anode, oxidation occurs. Solid $A$ loses $n$ electrons to become $A^{n+}$.

$$A(s) ightarrow A^{n+}(aq) + n e^-$$

At the cathode, reduction takes place. The $B^{2n+}$ ions need $2n$ electrons to deposit as solid $B$.

$$B^{2n+}(aq) + 2n e^- ightarrow B(s)$$

To balance the electron exchange, we must multiply the anode reaction by 2. This gives us our master overall reaction:

$$2A(s) + B^{2n+}(aq) ightarrow 2A^{n+}(aq) + B(s)$$

Now, let's construct the reaction quotient, $Q$. Remember, pure solids do not appear in this expression.

$$Q = \frac{[A^{n+}]^2}{[B^{2n+}]}$$

Substituting the given concentrations, we get:

$$Q = \frac{(2)^2}{1} = 4$$

### The Thermodynamic Bridge

Since the cell is at equilibrium ($E_{	ext{cell}} = 0$), the change in Gibbs free energy ($\Delta G$) is also zero.

We can bridge the gap between the current state and the standard state using the famous isotherm equation:

$$\Delta G = \Delta G^\circ + RT \ln Q$$

Setting $\Delta G$ to zero, we unlock the value of the standard Gibbs free energy:

$$0 = \Delta G^\circ + RT \ln(4)$$

$$\Delta G^\circ = -RT \ln(4) = -2RT \ln(2)$$

### The Enthalpy-Entropy Tango

Now, we bring in the heavy machinery of thermodynamics. The fundamental equation linking enthalpy, entropy, and free energy is:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

The problem gifts us a beautiful constraint: the standard enthalpy change is exactly twice the standard free energy change ($\Delta H^\circ = 2\Delta G^\circ$).

Let's substitute this into our fundamental equation:

$$\Delta G^\circ = 2\Delta G^\circ - T\Delta S^\circ$$

Rearranging this gives a remarkably simple relationship:

$$T\Delta S^\circ = \Delta G^\circ$$

$$\Delta S^\circ = \frac{\Delta G^\circ}{T}$$

### The Final Calculation and Physical Insight

We are at the finish line. Let's substitute our expression for $\Delta G^\circ$ into the entropy equation.

Notice how the temperature $T$ elegantly cancels out!

$$\Delta S^\circ = \frac{-2RT \ln(2)}{T} = -2R \ln(2)$$

Now, we just plug in the given values: $R = 8.3$ and $\ln(2) = 0.7$.

$$\Delta S^\circ = -2 	imes 8.3 	imes 0.7 = -11.62 	ext{ J K}^{-1} 	ext{mol}^{-1}$$

**The final answer is $-11.62$.**

But wait, let's pause and appreciate the physics here. The entropy change is negative, meaning the system became more ordered.

How is that possible when we went from 1 mole of aqueous ions to 2 moles?

This implies that the $A^{n+}$ ions have a much stronger hydration enthalpy. They tightly bind and order the surrounding water molecules, restricting their chaotic movement. The math perfectly captures this hidden microscopic reality!

Enter Numerical Value:

Visualized Solution

\text{Electrochemical Cell Setup}

\text{Balanced Cell Reaction}

\text{Reaction Quotient } (Q)

\text{Gibbs Free Energy and Nernst Equation}

\text{Calculating } \Delta G^\circ

\text{Thermodynamic Relation}

\text{Solving for } \Delta S^\circ

\text{Final Calculation}

\text{Conclusion \& Reflection}

The Sigma Insight: Electrochemical Cells

Decoding the Electrochemical Cell

The Chemical Choreography

The Thermodynamic Bridge

The Enthalpy-Entropy Tango

The Final Calculation and Physical Insight

Similar Questions

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