MCQ
Charge required to liberate  $11.5\,g $ sodium is
  • $0.5\,F$
  • B
    $0.1\,F$
  • C
    $1.5\,F$
  • D
    $96500$ coulombs

Answer

Correct option: A.
$0.5\,F$
a
(a) $N{a^ + } + {e^ - } \to Na$

Charge (in $F$)  $=$ moles of $e^-$  used $=$ moles of $Na$ deposited

$ = \frac{{11.5}}{{23}}\;gm = 0.5\,Faraday$.

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