Understanding the Problem

We are given an electrochemical cell: A(s) | Aⁿ⁺ (aq, 2 M) || B²ⁿ⁺ (aq, 1 M) | B(s). We are told that the standard enthalpy change (ΔH°) is twice the standard Gibbs free energy change (ΔG°) at 300 K. The cell's EMF is zero. We need to find the standard entropy change (ΔS°) per mole of B formed.

Key Concepts and Relationships

The fundamental equations connecting these thermodynamic quantities are:

1. The Gibbs Free Energy equation: $\Delta G°=\Delta H°-T\Delta S°$

2. The relationship between Gibbs Free Energy and Cell EMF: $\Delta G°=-nFE°$, where E° is the standard cell potential.

3. The Nernst Equation, which gives the cell EMF (E_cell) under non-standard conditions: $E_{\text{cell}}=E^{°}-\frac{RT}{nF}\ln Q$, where Q is the reaction quotient.

Step-by-Step Solution

Step 1: Find the Cell Reaction and 'n'

First, we deduce the balanced cell reaction. The cell notation is: A(s) | Aⁿ⁺(aq, 2 M) || B²ⁿ⁺(aq, 1 M) | B(s).

This means oxidation occurs at the left electrode (A(s) → Aⁿ⁺ + ne^-) and reduction occurs at the right electrode (B²ⁿ⁺ + 2e^- → B(s)).

To balance the electrons, we multiply the oxidation half-reaction by 2:
2A(s) → 2Aⁿ⁺ + 2ne^-
B²ⁿ⁺ + 2ne^- → B(s)

The overall cell reaction is:
$2A\left(s\right)+B^{2n+}\left(aq\right)\to 2A^{n+}\left(aq\right)+B\left(s\right)$

From this reaction, we see that n, the number of electrons transferred per mole of B formed, is 2.

Step 2: Apply the Given Relationship Between ΔH° and ΔG°

We are given: $\Delta H°=2\Delta G°$

Let's plug this into the Gibbs Free Energy equation:
$\Delta G°=\Delta H°-T\Delta S°$
$\Delta G°=(2\Delta G°)-T\Delta S°$

Now, let's solve for ΔS°:
$\Delta G°-2\Delta G°=-T\Delta S°$
$-\Delta G°=-T\Delta S°$
$\Delta G°=T\Delta S°$
This gives us a key relationship: $\Delta G°=T\Delta S°$ (Equation 1)

Step 3: Use the Condition that EMF is Zero

The cell EMF (E_cell) is given to be zero. We apply the Nernst equation:
$0=E^{°}-\frac{RT}{nF}\ln Q$
This simplifies to:
$E^{°}=\frac{RT}{nF}\ln Q$ (Equation 2)

Step 4: Find the Reaction Quotient (Q)

From our balanced reaction: 2A(s) + B²ⁿ⁺(aq) → 2Aⁿ⁺(aq) + B(s)
The reaction quotient Q is:
$Q=\frac{{\left[A^{n+}\right]}^2}{\left[B^{2n+}\right]}$
Substituting the given concentrations:
$Q=\frac{{\left(2\right)}^2}{1}=4$

Step 5: Relate E° to ΔG° and Solve

We know that $\Delta G°=-nF{E}^{°}$.
Therefore, $E^{°}=-\frac{\Delta G°}{nF}$.

Let's substitute this expression for E° into Equation 2:
$-\frac{\Delta G°}{nF}=\frac{RT}{nF}\ln Q$

We can cancel 'n' and 'F' from both sides (since they are non-zero):
$-\Delta G°=RT\ln Q$
$\Delta G°=-RT\ln Q$ (Equation 3)

Step 6: Combine Equations and Solve for ΔS°

We now have two expressions for ΔG°:
From Equation 1: $\Delta G°=T\Delta S°$
From Equation 3: $\Delta G°=-RT\ln Q$

Setting them equal to each other:
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