1. Write the cell reaction and identify electrons transferred

The reaction is $Zn(s) + Cu^{2+}(0.1\,M) \to Zn^{2+}(1\,M) + Cu(s)$

For this reaction:

• $Zn \to Zn^{2+} + 2e^-$
• $Cu^{2+} + 2e^- \to Cu$

So, the number of electrons transferred is $n=2$

2. Use the Nernst equation

At $298\,K$, $E_{cell}=E_{cell}^o-\frac{0.0591}{n}\log Q$

Given: $E_{cell}^o=1.10\,V$

3. Calculate the reaction quotient $Q$

For the reaction $Zn(s)+Cu^{2+}(aq)\to Zn^{2+}(aq)+Cu(s)$

Solids are not included in $Q$, so $Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$

Given concentrations: $[Zn^{2+}]=1\,M, \qquad [Cu^{2+}]=0.1\,M$

Hence, $Q=\frac{1}{0.1}=10$

4. Substitute into the Nernst equation

Since $\log 10 = 1$, $E_{cell}=1.10-\frac{0.0591}{2}(1)$

$E_{cell}=1.10-0.02955$

$E_{cell}=1.07045\,V$

Thus, $E_{cell}\approx 1.07\,V$

5. Check options

• A: $1.80\,V$ ❌
• B: $1.07\,V$ ✅
• C: $0.82\,V$ ❌
• D: $2.14\,V$ ❌

Therefore, the correct option is B.