Question

How many moles of electrons are required for the reduction of 1 mole of \( \mathrm{Cr^{3+}} \) to \( \mathrm{Cr^0} \) (s)?

• A. 1 mole of \( e^- \)
• B. 2 moles of \( e^- \)
• C. 3 moles of \( e^- \)
• D. None of these

Hint:

For reduction reactions, the number of electrons required is equal to the difference in oxidation states multiplied by the number of moles of the substance.

Answer 1

The Correct Option is C

Reduction involves the gain of electrons. The given reaction can be written as: \[ \mathrm{Cr^{3+} + 3e^- \rightarrow Cr^0}. \] From the reaction: Each \( \mathrm{Cr^{3+}} \) ion requires 3 electrons (\( 3e^- \)) to be reduced to \( \mathrm{Cr^0} \). Therefore, for 1 mole of \( \mathrm{Cr^{3+}} \), 3 moles of electrons are required. Key Points: The oxidation state of chromium changes from \( +3 \) in \( \mathrm{Cr^{3+}} \) to \( 0 \) in \( \mathrm{Cr^0} \). The number of moles of electrons required equals the change in oxidation state. Thus, the answer is \( \mathbf{3 \, \text{moles of} \, e^-} \).