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JEE Main 20204 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2020 · 4 September · Evening Shift · Chemistry · Electrochemistry

250mL of a waste solution obtained from the workshop of a goldsmith...

250mL250mL of a waste solution obtained from the workshop of a goldsmith contains 0.1MAgNO3and0.1MAuCl0 . 1M{AgNO}_{3} and 0 . 1MAuCl. The solution was electrolyzed at 2V2V by passing a current of 1A1A for 1515 minutes. The metal/metals electrodeposited will be : (EAg+/Ag0=0.80V,EAu+/Au0=1.69V)\left({\text{E}}_{{\text{Ag}}^{+} / \text{Ag}}^{0} = 0 . 80\text{V}, {\text{E}}_{{\text{Au}}^{+} / \text{Au}}^{0} = 1 . 69\text{V}\right)

Options

  1. Aonly gold
  2. Bsilver and gold in proportion to their atomic weights
  3. Conly silver
  4. Dsilver and gold in equal mass proportion

Correct answer

A. only gold

Step-by-step solution

Charge(q)=it96500F=1×15×6096500=90096500=9965F=0.0093F\text{Charge} ( q ) = \dfrac{\text{it}}{96500}F = \dfrac{1 \times 15 \times 60}{96500} = \dfrac{900}{96500} = \dfrac{9}{965}F = 0 . 0093F

No. of moles of Au+=0.025{Au}^{+} = 0 . 025 & No. of moles of Ag+=0.025{Ag}^{+} = 0 . 025

Species with higher value of SRPSRP will get deposited first at cathode.

(i)Au+(aqs)+e−⟶Au(s)( i ) {Au}^{+} ( aqs ) + {e}^{-} ⟶ Au ( s )

0.0250.00930 . 0250 . 0093 mole

so only AuAu will get deposited.

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