Question

An electric charge of $5$ Faradays is passed through three electrolytes $AgN{O}_3,CuS{O}_4$ and $FeC{l}_3$ solution. The grams of each metal liberated at cathode will be
[Atomic weight; $Fe-56g/mol,Cu-63.5g/mol,Ag-108g/mol]$

• A. Ag = 10.8 g, Cu = 12.7 g, Fe = 1.11 g
• B. Ag = 540 g, Cu = 367.5 g, Fe = 325 g
• C. Ag = 108 g, Cu = 63.5 g, Fe = 56 g
• D. Ag = 540 g, Cu = 158.8 g, Fe = 93.3 g

Answer 1

Answer: (D)

$A{g}^{+}+\underset{1\text{ mol }\equiv \text{ 1F}}{e^{-}}\to Ag$
$C{u}^{2+}+\underset{2\text{ mol }\equiv \text{ 2F}}{2e^{-}}\to Cu$
$F{e}^{3+}+\underset{3\text{ mol }\equiv \text{ 3F}}{3e^{-}}\to Fe$
Grams of Ag liberated $=\frac{108}{1}\times 5=540\text{ g}$
Grams of Cu liberated $=\frac{63.5}{2}\times 5=158.75\text{ g}$
Grams of Fe liberated $=\frac{56}{3}\times 5=93.3\text{ g}$