Question

How many moles of electrons are required to reduce 103.6 g of lead from $P{b}^{2+}$ to the metal?

• A. 0.5 mole
• B. 1 mole
• C. 2 moles
• D. 4 moles
• E. 8 moles

Answer 1

The correct option is B 1 mole

$P{b}^{2+}+2e^{-}\to Pb$

As, 2 moles of electrons are required to reduce 1 mole of $P{b}^{2+}$, therefore, for reduction of 103.6 gm $i.e$ 0.5 mole of $P{b}^{2+}$, only 1 mole of electrons will be required.