A) \[NaCl\]
B) \[BaC{{l}_{2}}\]
C) \[CuS{{O}_{4}}\]
D) \[AlC{{l}_{3}}\]
Correct Answer: A
Solution :
One Faraday deposits one equivalent which also one gram atom for \[Na\]. (i) \[Na_{(l)}^{+}+1{{e}^{-}}\xrightarrow{{}}Na(s);\](IF of electricity is required) \[Ba_{(l)}^{2+}+2{{e}^{-}}\xrightarrow{{}}Ba(s);\] (2F of electricity required) \[Cu_{(l)}^{2+}+2{{e}^{-}}\xrightarrow{{}}Cu(s);\] (2F of electricity is required) \[Al_{(l)}^{3+}+3{{e}^{-}}\xrightarrow{{}}Al(s);\] (3F of electricity is required)
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