Question:

The quantity of charge required to obtain 2 mol of Mn$^{2+}$ from MnO$_4^{-}$ is:

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In redox reactions, the number of moles of electrons transferred can be used to calculate the total charge required using Faraday’s constant (F).
Updated On: May 1, 2025
  • 2 F
  • 10 F
  • 5 F
  • 1 F
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The Correct Option is C

Solution and Explanation

To reduce MnO$_4^{-}$ to Mn$^{2+}$, the oxidation state of manganese changes from +7 in MnO$_4^{-}$ to +2 in Mn$^{2+}$, meaning 5 electrons are required for the reduction of 1 mol of MnO$_4^{-}$ to Mn$^{2+}$. For 2 mol of Mn$^{2+}$, 5 electrons are required for each mole, so the total number of moles of electrons required is: \[ \text{Electrons} = 5 \times 2 = 10 \text{ mol of electrons} \] To calculate the quantity of charge (Q), we use the relationship: \[ Q = n \times F \] Where: - \( n \) is the number of moles of electrons (10 mol), - \( F \) is Faraday’s constant, \( F = 96500 \, \text{C/mol} \).
Thus, the total charge required is: \[ Q = 10 \, \text{mol} \times 1 \, \text{F} = 10 \, \text{F} \]
Thus, the correct answer is 10 F.
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