Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • For the disproportionation reaction  <img alt="2Cu^{+}(aq) \rightleftharpoons Cu(s) + Cu^{2+}(aq) at , 298K," src="http://entrancecorner.oncodecogs.com/gif.latex?2Cu%5E%7B&plus;%7D%28aq%29%20%5Crightleftharpoons%20Cu%28s%29%20&plus;%2

For the disproportionation reaction  2Cu^{+}(aq) \rightleftharpoons Cu(s) + Cu^{2+}(aq) at , 298K,  lnK (where K is the equillibrium Constant )  is __________  \times 10^{-1}  Given \left ( E^{0}_{\frac{Cu^{2+}}{Cu+} } = 0.16V , E^{0}_{\frac{Cu^{+}}{Cu} } = 0.52V , \frac{RT}{F}=0.025\right )
 

Answers (1)

best_answer

We have given:

\mathrm{Cu^{2+}\: \overset{0.16V}{\rightarrow}\: Cu^{+}\: \overset{0.52}{\rightarrow}\: Cu}

Thus, Eodisproportionation = 0.52 - 0.16 = 0.36V

Now, we know:

\Delta G= -nFEo = -nRTlnK

\\\mathrm{\frac{RT}{F}\, lnK\: =\: 1\: x\: 0.36}\\\\\mathrm{lnK \: =\: \frac{0.36}{0.025}}

Thus, lnK = 14.4

Thus, lnK = 144 x 10-1

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions
anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel
anam-khan

CSAB 2025 counselling dates out for special, supernumerary rounds through JEE Mains

Latest Question