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) $N{a^ + } + {e^ - } \to Na$
(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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