A) \[1:1:1\]
B) \[1:2:3\]
C) \[3:2:1\]
D) \[6:3:2\]
E) \[2:3:1\]
Correct Answer: D
Solution :
The increasing order of deposition of cations at the cathode is \[C{{u}^{2+}}<A{{g}^{+}}<A{{u}^{3+}}\] \[E\propto Z\] \[A{{g}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Ag\] \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] \[A{{u}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Au\] 3 Faradays liberate 1 mole of Au, 3 moles of Ag and 3/2 moles of Cu. Thus molar ratio of \[Ag:Cu:Au\text{ }is\text{ }3:3/2:1\text{ }or\text{ }6:3:2\].
You need to login to perform this action.
You will be redirected in
3 sec
