Question

An electric charge of 5 Faradays is passed through three electrolytes $AgNO_{3},CuSO_{4}$ and $FeCl_{3}$solution the grams of each metal liberated at cathode will be

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

Answer 1

Answer: (D)

$Ag = 540g,Cu = 158.8g,Fe = 93.3g$

Answer 2

$Ag^{+} + \underset{1mol \equiv 1F}{e^{-}} \rightarrow Ag$

$Cu^{2 +} + \underset{2mol \equiv 2F}{2e^{-}} \rightarrow Cu$

$Fe^{3 +} + \underset{3mol \equiv 3F}{3e^{-}} \rightarrow Fe$

Grams of Ag liberated $= \frac{108}{1} \times 5 = 540g$

Grams of Cu liberated $= \frac{63.5}{2} \times 5 = 158.75g$

Grams of Fe liberated $= \frac{56}{3} \times 5 = 93.3g$