A) 0.5 F
B) 0.1 F
C) 1.5 F
D) 96500 coulombs
Correct Answer: 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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