For the following reaction, at equilibrium the Nernst equation may be written as:

$Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)$

$E_{cell}^0=\frac{2.303RT}{2F}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$

The correct answer is Option (2) → $E_{cell}^0=\frac{2.303RT}{2F}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$ 

We derive the correct Nernst equation for the reaction:

$Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)$

Step 1: Write the general Nernst equation 

$E_{cell}= E_{cell}^0-\frac{2.303RT}{nF}\log Q$

Where

• n = number of electrons transferred

• Q = reaction quotient

Step 2: Find number of electrons (n)

Half reactions:

Oxidation: $Zn → Zn^{2+} + 2e^-$

Reduction: $Cu^{2+} + 2e^- →Cu$

So,

$n = 2$

Step 3: Write reaction quotient Q

Pure solids are not included in Q.

$Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$

Step 4: Substitute into Nernst equation

$E_{cell}= E_{cell}^0-\frac{2.303RT}{nF}\log (\frac{[Zn^{2+}]}{[Cu^{2+}]})$

Step 5: At equilibrium

At equilibrium:

$E_{cell}=0$

So,

$E_{cell}^0=\frac{2.303RT}{nF}\log (\frac{[Zn^{2+}]}{[Cu^{2+}]})$2F

Final Answer (correct option)

$E_{cell}^0=\frac{2.303RT}{2F}\log\frac{[Zn^{2+}]}{[Cu^{2+}]}$