How many moles of electrons are required for the reduction of 1 mole of \( \mathrm{Cr^{3+}} \) to \( \mathrm{Cr^0} \) (s)?
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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.
Reduction is defined as the acceptance of electrons. The provided reaction is depicted as:\[\mathrm{Cr^{3+} + 3e^- \rightarrow Cr^0}.\]Based on this reaction: It takes 3 electrons (\( 3e^- \)) for each \( \mathrm{Cr^{3+}} \) ion to achieve reduction to \( \mathrm{Cr^0} \). Consequently, 3 moles of electrons are necessary for 1 mole of \( \mathrm{Cr^{3+}} \).Key Information: The oxidation state of chromium transitions from \( +3 \) in \( \mathrm{Cr^{3+}} \) to \( 0 \) in \( \mathrm{Cr^0} \). The quantity of moles of electrons needed corresponds directly to the magnitude of the change in oxidation state.Hence, the required quantity is \( \mathbf{3 \, \text{moles of} \, e^-} \).
