If the number of moles of \(\text{Fe}^{2+}\) ions oxidized by one mole of acidified \(\text{MnO}_4^-\) is \(x\), the number of moles of \(\text{Fe}^{2+}\) ions oxidized by one mole of acidified \(\text{Cr}_2\text{O}_7^{2-}\) is
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Equivalents of oxidizing agent = Equivalents of reducing agent.
\(\text{Moles} \times \text{n-factor} = \text{Moles} \times \text{n-factor}\).
For \(\text{MnO}_4^-\), n-factor=5. For \(\text{Cr}_2\text{O}_7^{2-}\), n-factor=6.
Step 1: Determine \(x\) for \(\text{MnO}_4^-\)
In acidic medium, \(\text{MnO}_4^-\) is reduced to \(\text{Mn}^{2+}\).
Change in oxidation state of Mn: \(+7 \to +2\).
n-factor for \(\text{MnO}_4^- = 5\).
Reaction: \(\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\).
Thus, 1 mole of \(\text{MnO}_4^-\) oxidizes 5 moles of \(\text{Fe}^{2+}\).
So, \(x = 5\).
Step 2: Determine moles for \(\text{Cr}_2\text{O}_7^{2-}\)
In acidic medium, \(\text{Cr}_2\text{O}_7^{2-}\) is reduced to \(\text{Cr}^{3+}\).
Change in oxidation state of Cr: \(+6 \to +3\). Since there are 2 Cr atoms, total change = \(2 \times 3 = 6\).
n-factor for \(\text{Cr}_2\text{O}_7^{2-} = 6\).
Reaction: \(\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \to 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\).
Thus, 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\) oxidizes 6 moles of \(\text{Fe}^{2+}\).
Step 3: Express the result in terms of \(x\)
We need the value 6.
We have \(x = 5\).
Substitute \(x=5\) into the options:
(A) \(5(5)/8 = 25/8\)
(B) \(6(5)/5 = 6\) (Matches)
(C) \(8(5)/5 = 8\)
(D) \(5(5)/6 = 25/6\)
Final Answer: \(\frac{6x}{5}\).
