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.