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  4. In passing 3 faraday of electricity through the three electrolytic cells connected in series containing Ag+ , Cu2+ and Al+3 respectively. The molar ratio in which the three metal ions are liberated at the electrode is

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Q. In passing $3$ faraday of electricity through the three electrolytic cells connected in series containing $Ag^+ , Cu^{2+}\,$ and $\,Al^{+3}$ respectively. The molar ratio in which the three metal ions are liberated at the electrode is

Electrochemistry

Solution:

$Ag^+ + e^- \rightarrow Ag$
$Ca^{2+} + 2e^- \rightarrow Ca$
$Al^{3+} + 3e^- \rightarrow Al$
$3$ faradays liberate $1$ mol of $Al$, $3$ moles of $Ag$ and $3/2$ moles of $Cu$. Thus, molar ratio of
$Ag : Cu : Al$ is $3 : 3/2 : 1$ or $6 : 3 : 2$