250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1M AgNO3 and 0.1M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be : (E°Ag+/Ag = 0.8 V, E°Au+/Au = 1.69 V)

Option 1 - <p>Silver and gold in equal mass proportion<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>Only silver<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>Only gold<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>Silver and gold in proportion to their atomic weights<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;<br>&lt;!--[endif]--&gt;</p>
9 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 3
Detailed Solution:

Charge (q) = (it/96500) F = (1x15x60)/96500 = 900/96500 = 9/965 F = 0.0093F
No. of moles of Au? = 0.025 & No. of moles of Ag? = 0.025
Species with higher value of SRP will get deposited first at cathode.
(i) Au? (aqs) + e? → Au (s)
0.025 0.0093 mole
So only Au will get deposited

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ΔG° = –RT * 2.303 log K

–nFE° = +RT * 2.303 log K

2 * 96500 * 0.295 = 8.314 * 298 * 2.303 log10 K

10 = log10 K = 1010

It has chiral centre and differently di substituted double bonded carbon atoms.

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Chemistry Electrochemistry 2025

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