JEE Main 2020 — Chemistry Physical Chemistry

2020mcqmedium

$250\mathrm{mL}$ of a waste solution obtained from the workshop of a goldsmith contains $0.1M{\mathrm{AgNO}}_{3}$ and $0.1M\mathrm{AuCl}$ . The solution was electrolyzed at $2V$ by passing a current of $1A$ for $15$ minutes. The metal/metals electrodeposited will be : $({\text{E}}_{{\text{Ag}}^{+}/\text{Ag}}^{0}=0.80\text{V},{\text{E}}_{{\text{Au}}^{+}/\text{Au}}^{0}=1.69\text{V})$

Official previous-year question

Held on 4 Sept 2020 · Verified 6 Jul 2026.

Options

A
only gold
B
silver and gold in proportion to their atomic weights
C
only silver
D
silver and gold in equal mass proportion

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Solution

$\text{ Charge }(q)=\frac{\text{ it }}{96500}F=\frac{1\times 15\times 60}{96500}=\frac{900}{96500}=\frac{9}{965}F=0.0093F$

No. of moles of ${\mathrm{Au}}^{+}=0.025$ & No. of moles of ${\mathrm{Ag}}^{+}=0.025$

Species with higher value of $\mathrm{SRP}$ will get deposited first at cathode.

$(i){\mathrm{Au}}^{+}(\mathrm{aqs})+{e}^{-}\longrightarrow \mathrm{Au}(s)$

$0.0250.0093$ mole

so only $\mathrm{Au}$ will get deposited.

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Solution

$\text{ Charge }(q)=\frac{\text{ it }}{96500}F=\frac{1\times 15\times 60}{96500}=\frac{900}{96500}=\frac{9}{965}F=0.0093F$

No. of moles of ${\mathrm{Au}}^{+}=0.025$ & No. of moles of ${\mathrm{Ag}}^{+}=0.025$

Species with higher value of $\mathrm{SRP}$ will get deposited first at cathode.

$(i){\mathrm{Au}}^{+}(\mathrm{aqs})+{e}^{-}\longrightarrow \mathrm{Au}(s)$

$0.0250.0093$ mole

so only $\mathrm{Au}$ will get deposited.

JEE Main 2020 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2020 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry