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Step-by-Step Solution
Step 1: Identify the Half-Reactions
• Oxidation (at the Zinc electrode):
Zn \rightarrow Zn^{2+} + 2\,e^- .
• Reduction (at the Iron electrode):
2\,Fe^{3+} + 2\,e^- \rightarrow 2\,Fe^{2+} .
Step 2: Write the Overall Reaction
Combining the two half-reactions gives:
Zn + 2\,Fe^{3+} \rightarrow Zn^{2+} + 2\,Fe^{2+} .
Step 3: Calculate the Standard Cell Potential
Using the standard reduction potentials provided:
E^0_{(Fe^{3+}/Fe^{2+})} = 0.77\,V and E^0_{(Zn^{2+}/Zn)} = -0.76\,V .
The standard cell potential is:
E^0_{\text{cell}} = 0.77\,V - (-0.76\,V) = 0.77\,V + 0.76\,V = 1.53\,V .
Step 4: Apply the Nernst Equation
At 25°C (approximately), the Nernst equation for this cell can be written as:
E_{\text{cell}} = E^0_{\text{cell}} - \dfrac{0.06}{n} \,\log Q,
where n = 2 (number of electrons transferred) and
Q = \dfrac{[Zn^{2+}]\,[Fe^{2+}]^2}{[Fe^{3+}]^2} .
Because [Zn^{2+}] is 1 M, we simplify the relevant ratio to
Q = \dfrac{[Fe^{2+}]^2}{[Fe^{3+}]^2} .
We are told the cell potential is 1.50\,V . Thus:
1.50 = 1.53 - \dfrac{0.06}{2}\,\log\!\biggl(\dfrac{[Fe^{2+}]}{[Fe^{3+}]}\biggr)^2.
Step 5: Solve for the Concentration Ratio
Rearrange:
1.53 - 1.50 = 0.03 = \dfrac{0.06}{2} \,\log\!\bigl(\dfrac{[Fe^{2+}]}{[Fe^{3+}]}\bigr)^2.
Hence,
\log\!\bigl(\dfrac{[Fe^{2+}]}{[Fe^{3+}]}\bigr)^2 = \dfrac{0.03}{0.06/2}
= \dfrac{0.03}{0.03} = 1.
So,
\bigl(\dfrac{[Fe^{2+}]}{[Fe^{3+}]}\bigr)^2 = 10
\quad\Longrightarrow\quad \dfrac{[Fe^{2+}]}{[Fe^{3+}]} = \sqrt{10}\approx 3.16.
Step 6: Find the Fraction of Total Iron Present as Fe3+
Let [Fe^{3+}] = a and [Fe^{2+}] = b . From the ratio b/a = \sqrt{10} , we have b = \sqrt{10}\,a .
The total iron concentration is a + b = a + \sqrt{10}\,a = a(1 + \sqrt{10}).
So, the fraction of total iron as Fe^{3+} is:
\dfrac{a}{a + \sqrt{10}\,a}
= \dfrac{1}{1 + \sqrt{10}}.
Numerically,
\dfrac{1}{1 + \sqrt{10}} \approx \dfrac{1}{1 + 3.16} \approx \dfrac{1}{4.16} \approx 0.24.
This fraction is 0.24 = 24 \times 10^{-2}.
Hence, the value of x (nearest integer) is 24.
