1. Identify the half-cells

The cell is:

$\text{Zn} \mid \text{Zn}^{2+}(1\text{ M}) \parallel \text{Fe}^{3+},\text{Fe}^{2+} \mid \text{Pt}$

Given standard reduction potentials:

$E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77\text{ V}$ $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$

So, under standard conditions,

• Cathode: $\text{Fe}^{3+} + e^- \to \text{Fe}^{2+}$
• Anode: $\text{Zn} \to \text{Zn}^{2+} + 2e^-$

2. Calculate standard cell potential

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

$E^\circ_{\text{cell}} = 0.77 - (-0.76) = 1.53\text{ V}$

3. Write the overall balanced reaction

Oxidation:

$\text{Zn} \to \text{Zn}^{2+} + 2e^-$

Reduction:

$2\text{Fe}^{3+} + 2e^- \to 2\text{Fe}^{2+}$

Overall:

$\text{Zn} + 2\text{Fe}^{3+} \to \text{Zn}^{2+} + 2\text{Fe}^{2+}$

Thus, $n=2$ electrons.

4. Apply Nernst equation

At $25^\circ$C,

$E = E^\circ - \frac{0.0591}{n} \log Q$

For the reaction,

$Q = \frac{[\text{Zn}^{2+}][\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}$

Since $[\text{Zn}^{2+}] = 1\text{ M}$,

$Q = \frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}$

So,

$E = 1.53 - \frac{0.0591}{2} \log\left(\frac{[\text{Fe}^{2+}]^2}{[\text{Fe}^{3+}]^2}\right)$

This simplifies to

$E = 1.53 - 0.0591 \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right)$

Given $E = 1.500\text{ V}$,

$1.500 = 1.53 - 0.0591 \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right)$

$0.0591 \log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right) = 1.53 - 1.50 = 0.03$

$\log\left(\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right) = \frac{0.03}{0.0591} \approx 0.508$

$\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 10^{0.508} \approx 3.22$

5. Find fraction of total iron present as $\text{Fe}^{3+}$

Let total iron concentration be

$[\text{Fe}^{3+}] + [\text{Fe}^{2+}]$

Fraction present as $\text{Fe}^{3+}$ is

$f = \frac{[\text{Fe}^{3+}]}{[\text{Fe}^{3+}] + [\text{Fe}^{2+}]}$

Using

$\frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} = 3.22$

Let $[\text{Fe}^{3+}] = a$, then $[\text{Fe}^{2+}] = 3.22a$.

So,

$f = \frac{a}{a+3.22a} = \frac{1}{4.22} \approx 0.237$

Thus,

$f \approx 23.7 \times 10^{-2}$

Hence,

$x \approx 24$

6. Final answer

The nearest integer value of $x$ is:

$\boxed{24}$