1. Cell reaction

For the cell $Zn(s)\,|\,ZnSO_4(aq)\,||\,CuSO_4(aq)\,|\,Cu(s)$

The overall reaction is: $Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)$

Here, number of electrons transferred: $n=2$

2. Reaction quotient

For the reaction, $Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$

Given: $[Zn^{2+}] = 10[Cu^{2+}]$ So, $Q=10$

3. Use Nernst equation

At general temperature $T$, $E=E^\circ-\frac{RT}{nF}\ln Q$

Substitute $E^\circ=1.1\,V$, $n=2$, and $Q=10$: $E=1.1-\frac{RT}{2F}\ln 10$

Since $\ln 10 = 2.303$ we get $E=1.1-\frac{2.303RT}{2F}$

4. Relate $\Delta G$ and cell emf

We use: $\Delta G=-nFE$

So, $\Delta G=-2F\left(1.1-\frac{2.303RT}{2F}\right)$

Now simplify: $\Delta G=-2.2F+2.303RT$

Thus, $\boxed{\Delta G=2.303RT-2.2F}$

5. Option check

• A: $1.1F$ ❌
• B: $2.303RT-2.2F$ ✅
• C: $2.303RT+1.1F$ ❌
• D: $-2.2F$ ❌

Hence the correct option is B.