What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the reaction below to obtain chromium metal
Cr₂O₃(s) + 2Al(s) $\rarr$ 2Cr(l) + Al₂O₃(s)

Question

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the reaction below to obtain chromium metal
Cr₂O₃(s) + 2Al(s)

\rarr

2Cr(l) + Al₂O₃(s)

Quizplus · Accepted Answer

The reaction involves the reduction of chromium(III) oxide (Cr2O3) to chromium metal (Cr) and the oxidation of aluminum (Al) to aluminum oxide (Al2O3). Each Cr3+ ion gains 3 electrons to become Cr, and there are 2 Cr atoms in Cr2O3, so a total of 6 electrons are gained by chromium. Since the number of electrons lost by the reducing agent (Al) must equal the number gained by the oxidizing agent (Cr2O3), 6 moles of electrons are exchanged in total.

What Is the Total Number of Moles (N) of Electrons →\rarr→

What Is the Total Number of Moles (N) of Electrons $\rarr$