Consider the cell at 25$$^\circ$$C
Zn|Zn$$^{2+}$$ aq, 1 M || Fe$$^{3+}$$(aq), Fe$$^{2+}$$aq|Pt
The fraction of total iron present as Fe$$^{3+}$$ ion at the cell potential of 1.500 V is $$x \times 10^{-2}$$. The value of x is ___.
(Nearest integer)
Given: E$$^\circ_{Fe^{3+}|Fe^{2+}}$$ = 0.77V, E$$^\circ_{Zn^{2+}|Zn}$$ = -0.76V

We first list the standard reduction potentials at 25 °C:

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

The given cell is

$$$\text{Zn}\;|\;\text{Zn}^{2+}(1\ \text{M})\;||\;\text{Fe}^{3+}(aq),\, \text{Fe}^{2+}(aq)\;|\;\text{Pt}.$$$

On the left, zinc is oxidised, and on the right, iron(III) is reduced. Writing the two half-reactions with the electrons explicitly, we have

Oxidation (anode): $$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-.$$

Reduction (cathode): $$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}.$$

To balance electrons, we multiply the iron half-reaction by 2:

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

Adding the two balanced half-reactions gives the overall cell reaction,

$$$\text{Zn} + 2\,\text{Fe}^{3+} \rightarrow \text{Zn}^{2+} + 2\,\text{Fe}^{2+}.$$$

For the balanced reaction, the number of electrons transferred is $$n = 2.$$

Now we calculate the standard cell potential. The formula is

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

Substituting the given values,

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

The measured cell potential at 25 °C is $$E_{\text{cell}} = 1.500\ \text{V}.$$ The zinc compartment is already at standard conditions ($$[\text{Zn}^{2+}] = 1\ \text{M}$$), so any departure from the standard potential must arise from the iron couple. We therefore write the Nernst equation for the entire cell.

The general Nernst equation at 25 °C is

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \dfrac{0.0591}{n}\,\log Q,$$

where $$Q$$ is the reaction quotient. From the balanced reaction,

$$Q = \dfrac{[\text{Zn}^{2+}]\, [\text{Fe}^{2+}]^{2}}{[\text{Fe}^{3+}]^{2}}.$$

Because $$[\text{Zn}^{2+}] = 1\ \text{M},$$ this simplifies to

$$Q = \left(\dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\right)^{2}.$$

Let $$R = \dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}.$$ Then $$Q = R^{2}.$$

We now insert the numerical values into the Nernst equation:

$$1.500 = 1.53 - \dfrac{0.0591}{2}\,\log(R^{2}).$$

Simplifying, the coefficient of the logarithm is

$$\dfrac{0.0591}{2} = 0.02955.$$

Rewriting the equation,

$$$0.02955\,\log(R^{2}) = 1.53 - 1.500 = 0.030.$$$

Dividing both sides by 0.02955, we get

$$\log(R^{2}) = \dfrac{0.030}{0.02955} \approx 1.015.$$

Taking antilogarithms,

$$R^{2} = 10^{1.015} \approx 10.36.$$

Hence

$$R = \sqrt{10.36} \approx 3.22.$$

This value means

$$\dfrac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]} \approx 3.22.$$

For convenience let the total iron concentration be

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

Because $$[\text{Fe}^{2+}] = 3.22[\text{Fe}^{3+}],$$ we write

$$$C = 3.22[\text{Fe}^{3+}] + [\text{Fe}^{3+}] = 4.22[\text{Fe}^{3+}].$$$

The fraction of iron that is present as $$\text{Fe}^{3+}$$ is therefore

$$$\text{Fraction} = \dfrac{[\text{Fe}^{3+}]}{C} = \dfrac{[\text{Fe}^{3+}]}{4.22[\text{Fe}^{3+}]} = \dfrac{1}{4.22} \approx 0.237.$$$

Expressed in the required form,

$$0.237 \;=\; 23.7 \times 10^{-2}.$$

Rounded to the nearest integer, $$x = 24.$$

So, the answer is $$24$$.