A) 2F, 6F, 3F, and 1F, respectively
B) 6F, 2F, 3F and 1F, respectively
C) 2F, 6F, 1F and 3F, respectively
D) 6F, 2F, 1F and 3F, respectively
Correct Answer: A
Solution :
\[CuS{{O}_{4}}+2{{e}^{-}}\xrightarrow{{}}Cu+SO_{4}^{\,-}\] \[B{{i}_{2}}{{(S{{O}_{4}})}_{3}}+6{{e}^{-}}\xrightarrow{{}}2Bi+3SO_{4}^{-}\] \[AlC{{l}_{3}}+3{{e}^{-}}\xrightarrow{{}}Al+3C{{l}^{-}}\] \[AgN{{O}_{3}}+{{e}^{-}}\xrightarrow{{}}Ag+NO_{3}^{-}\] In 1 M\[CuS{{O}_{4}}\] solution, 1 M\[B{{i}_{2}}{{(S{{O}_{4}})}_{3}}\]solution, 1 M\[AlC{{l}_{3}}\] solution and 1 M\[AgN{{O}_{3}}\]solution, 2 mole electron, 6 mole electron, 3 mole electron and 1 mole electron are needed to deposit Cu, Bi, Al and Ag at the cathode, respectively. But one mole electron = 1 F electricity That's why number of Faraday required to deposit 1 M of each \[CuS{{O}_{4}},\] \[B{{i}_{2}}{{(S{{O}_{4}})}_{3}},\] and \[AgN{{O}_{3}}\] solution are 2F, 6F, 3P and F, respectively.
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