A) 0.005 mol
B) 0.01 mol
C) 0.02 mol
D) 0.04 mol
Correct Answer: C
Solution :
\[{{[Cu{{(C{{H}_{3}}CN)}_{4}}]}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Cu+4C{{H}_{3}}CN\] For the above reaction 1 F charge is required to deposit 1 mole of Cu (i.e. 63.5 g) Charge supplied \[=0.965\times 2000\] \[=2\times 965=0.02\text{ }F\] \[\therefore \]Moles of Cu deposited = 0.02 molYou need to login to perform this action.
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