3pb2+(aq) + 2cr(s) → 3pb(s) + 2cr3+(aq) what is the number of moles of electrons gained by 3.0 moles of lead ions?

To determine the electrons gained in this redox reaction, we must balance the charges on the side containing the reduced substance and the side with oxidized substance. In this case, the reduced substance is Pb2+ and the oxidized substance is Cr. In each side, 6 electrons are gained or lose to balance the charges. The answer is 6 electrons.

6 moles of Electrons

Explanation: balanced equation: 3 Pb+2(aq) + 2 Cr(s)> 3 Pb(s) + 2 Cr+3(aq)

Here oxidation no of Cr changes from 0 to +3 hence total no of electrons lost by Cr are 6 since one mole of Cr =3 Electrons lost ,and 2 moles of Cr =2x3=6 moles of Electrons.

These 6 moles of electrons lost by Cr are gained by 3 moles of Pb+2 and gets reduced into Pb.

So total 6 moles of Electrons are involved in Redox reaction

False I would say because it’s different