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  • 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 )
 

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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

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