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`        Question Number 23 from electrochemistry....................... ... `
one year ago

Sahani Kumar
99 Points
```							Option (1)  is correct.    I have explained it below.  Given:E cell = 1.10 volt.$E^o _c_e_l_l$ = – (0.059/2)logKc.  Kc is a equilibrium constantNow on putting the value of E cell in above equation we get 1.10= -- (0.059/2)logKc–( 1.10 * 2)/0.059 = log Kc.    on solving we will get– 37=logKcSo Kc=10^-37RegardsAskiitian member If you like this answer please encourage us by approving this answer.
```
one year ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions