Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
Calculate ΔGo for the following cell reaction
Zn s + Ag 2 O s + H 2 O l Zn 2+ aq +2Ag s +2 OH aq

EAg+/Ag0=+0.80V and EZn+2/Zn0=-0.76V ,F=96500

(Given:  K sp ofAgOH=2× 10 8 )
ChemistryElectrochemistryJEE Main
Options:
  • A -305kJ/mol
  • B 212 kJ/mol
  • C 305 kJ/mol
  • D 301 kJ/mol
Solution:
2688 Upvotes Verified Answer
The correct answer is: 212 kJ/mol
Calculation for EOH-Ag2OAg-
Reduction [2Ag++2e-2Ags,G1o=-2×F×EoAg+|Ag
Ionisation [Ag2O+H2Ol2Ag++2OH-G2o=-2.303RTlogksp2
Net Ag2O+H2Ol+2e- 2Ags+2OH-,G3o=-2×F×EOH-Ag2OAgo
G3o=G1o+G2o
+2×F×EOH-Ag2OAgo=+2×F×EoAg+|Ag+2.303RTlogksp22F
EOH-Ag2OAgo=EoAg+|Ag+0.062×2logksp
=0.8+0.06log2×10-8
=0.8+0.06-8+0.3
=0.8+0.06×-7.7
=0.8-0.462
=0.338Volt0.34V Go=-nFEceuo
=-nFEco-EARed
=-2×EOH-Ag2OAgo-EZn+2|Zno
=-2×96500×0.34+0.76
=-2×96500×1.1volt
=212300J
=212.3 kJ

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.