Step-by-step Solution:

1. Determine the cell reaction and the number of electrons transferred.

The electrochemical cell is given as: $A\left( s \right)\left| {{A^{n + }}\left( {aq,2M} \right)} \right|{B^{2n + }}\left( {aq,1M} \right)\left| {B\left( s \right).} \right.$

• Anode (Oxidation): At the anode, metal A is oxidized to its ion $A^{n+}$. $A(s) \to A^{n+}(aq) + ne^-$

• Cathode (Reduction): At the cathode, ion $B^{2n+}$ is reduced to metal B. $B^{2n+}(aq) + 2ne^- \to B(s)$

• Overall Cell Reaction: To balance the electrons, we multiply the anode reaction by 2. $2A(s) \to 2A^{n+}(aq) + 2ne^-$ $B^{2n+}(aq) + 2ne^- \to B(s)$ Adding the two half-reactions gives the overall reaction: $2A(s) + B^{2n+}(aq) \to 2A^{n+}(aq) + B(s)$

The number of moles of electrons transferred in this balanced reaction is $z = 2n$.

2. Apply the Nernst Equation.

The Nernst equation relates the cell potential ($E_{cell}$) to the standard cell potential ($E^\circ_{cell}$) and the reaction quotient ($Q$). $E_{cell} = E^\circ_{cell} - \frac{RT}{zF} \ln Q$

The reaction quotient $Q$ for this reaction is: Q = \frac{{\left[ {{A^{n + }}} \right]^2}}}{{\left[ {{B^{2n + }}} \right]}} Given concentrations are $[A^{n+}] = 2M$ and $[B^{2n+}] = 1M$. $Q = \frac{{(2)^2}}{1} = 4$

Substituting $z = 2n$ and $Q = 4$ into the Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{RT}{2nF} \ln(4)$

3. Use the given condition that the EMF of the cell is zero.

We are given that $E_{cell} = 0$. $0 = E^\circ_{cell} - \frac{RT}{2nF} \ln(4)$ $E^\circ_{cell} = \frac{RT}{2nF} \ln(4) = \frac{RT}{2nF} (2\ln(2))$ $E^\circ_{cell} = \frac{RT}{nF} \ln(2)$

4. Calculate the standard Gibbs free energy change ($ΔG^\circ$).

The relationship between $ΔG^\circ$ and $E^\circ_{cell}$ is: $ΔG^\circ = -zFE^\circ_{cell}$ Substituting $z = 2n$ and the expression for $E^\circ_{cell}$: $ΔG^\circ = -(2n)F \left( \frac{RT}{nF} \ln(2) \right)$ $ΔG^\circ = -2RT \ln(2)$

5. Use the thermodynamic relationship between $ΔG^\circ$, $ΔH^\circ$, and $ΔS^\circ$.

The Gibbs-Helmholtz equation is: $ΔG^\circ = ΔH^\circ - TΔS^\circ$

We are given that $ΔH^\circ = 2ΔG^\circ$ at $T = 300K$. Substituting this into the equation: $ΔG^\circ = 2ΔG^\circ - TΔS^\circ$ $-ΔG^\circ = -TΔS^\circ$ $ΔS^\circ = \frac{ΔG^\circ}{T}$

6. Calculate the value of $ΔS^\circ$.

Substitute the expression for $ΔG^\circ$ from step 4 into the equation for $ΔS^\circ$: $ΔS^\circ = \frac{-2RT \ln(2)}{T}$ $ΔS^\circ = -2R \ln(2)$

Now, plug in the given values: $R = 8.3 J\,K^{-1}\,mol^{-1}$ and $\ln(2) = 0.7$. $ΔS^\circ = -2 \times (8.3 J\,K^{-1}\,mol^{-1}) \times 0.7$ $ΔS^\circ = -16.6 \times 0.7$ $ΔS^\circ = -11.62 J\,K^{-1}\,mol^{-1}$

The calculated value of $ΔS^\circ$ corresponds to the overall cell reaction, which involves the formation of one mole of B. Therefore, the standard entropy change per mole of B formed is $-11.62 J\,K^{-1}\,mol^{-1}$.