To solve this redox equation, we divide the equation into two parts: the oxidation and the reduction part. The reduction part is 3 Pb 2+ = 3 Pb while the oxidation part is 2 Cr = 2 Cr3+. We balance the charges. 3 Pb 2+ + 6e- = 2 Pb and 2 Cr = 2 Cr 3+ + 6e-. Adding the two equations, the 6 e- just cancels out. Hence for 3 moles of lead ions, there are 6 moles of electrons gained.
