Question

Charge required to liberate $11.5g$ sodium is

• A. $96500C$
• B. $0.1F$
• C. $1.5F$
• D. $0.5F$

Answer 1

Answer: (D)

$N{a}^{+}+e^{-}\to Na$
Charge (in F) = moles of $e^{-}$ used = moles of Na deposited =$\frac{11.5}{25}g=0.5F$