Question

solid al is reduced from al(no₃)₃ solution. how many moles of electrons are transferred during the reduction of al?

Explanation:

Step1: Determine oxidation state of Al in \( \text{Al(NO}_3\text{)}_3 \)

In \( \text{Al(NO}_3\text{)}_3 \), nitrate (\( \text{NO}_3^- \)) has a charge of -1. Let the oxidation state of Al be \( x \). The compound is neutral, so \( x + 3\times(-1) = 0 \). Solving, \( x = +3 \).

Step2: Determine reduction half - reaction

The reduction of \( \text{Al}^{3+} \) (from \( \text{Al(NO}_3\text{)}_3 \)) to solid Al (\( \text{Al}^0 \)) is \( \text{Al}^{3+}+3e^-
ightarrow\text{Al} \). For 1 mole of Al atoms formed, 3 moles of electrons are gained (transferred) as the oxidation state of Al changes from +3 to 0, requiring 3 electrons per Al ion.

Answer:

For the reduction of 1 mole of Al from \( \text{Al(NO}_3\text{)}_3 \) solution, 3 moles of electrons are transferred.