Question

How many moles of electrons are transferred in the following reaction?
$C{e}^{3+}+Pb\to Ce+P{b}^{4+}$

• A. 14
• B. 12
• C. 7
• D. 24
• E. 3

Answer 1

Answer: (B)

12

The balanced redox reaction is
$C{e}^{3+}+Pb\to Ce+P{b}^{4+}$.
The oxidation half reaction is
$Pb\to P{b}^{4+}+4e^{-}$.
The reduction half reaction is
$C{e}^{3+}+3e^{-}\to Ce$ .
To balance the number of electrons, multiply oxidation half reaction with 3 and the reduction half reaction with 4.
The oxidation half reaction is
$3Pb\to 3P{b}^{4+}+12e^{-}$.
The reduction half reaction is
$4C{e}^{3+}+12e^{-}\to 4Ce$.
Add oxidation half reaction and reduction half reaction.
$4C{e}^{3+}+3Pb\to 4Ce+3P{b}^{4+}$
The number of electrons transferred is 12.