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  • Given that the equilibrium constant for the reaction  <img alt="\mathrm{Z n(s)+Sn^{2+}(a q)\rightleftharpoons Zn^{2+}(a q)+S n(s)}" src="/latex-image/?%5Cmathrm%7BZ%20n%28s%29&plus;Sn

Given that the equilibrium constant for the reaction  \mathrm{Z n(s)+Sn^{2+}(a q)\rightleftharpoons Zn^{2+}(a q)+S n(s)}  is  \mathrm{1 \times 10^{16}}  at 298 K. The magnitude of standard potential of \mathrm{\mathrm{Sn} / \mathrm{Sn}^{2+}}  if  \mathrm{ E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V}}  is  {Take\frac{2.303 \mathrm{RT}}{\mathrm{~F}} =0.05 9 \mathrm{~V}}

Option: 1

19 \times 10^{-1}


Option: 2

15 \times 10^{-2}


Option: 3

29 \times 10^{-2}


Option: 4

32 \times 10^{-2}


Answers (1)

\mathrm{\mathrm{Zn}(s)+Sn^{2+}(a q) \rightleftharpoons Z n^{+2}(a q)+Sn(s) }
\mathrm{E_{\text {cell }}^0=\frac{2.303 R T}{2 F} \log k }
\mathrm{E_{\text {cell }}^0=\frac{0.059}{2} \log \left(10^{16}\right) \\ }
\mathrm{E_{\text {cell }}^0=\frac{0.059}{2} \times 16 \\ }
\mathrm{E_{\text {cell }}^00.4 7 \mathrm{~V} \\ }
\mathrm{E_{\text {cell }}^0 =E^o _{\mathrm{Sn}^{2+} / S n}-E^o_{Z n^{2+} / \mathrm{Zn}} }
\mathrm{E_{\text {cell }}^0 {{\mathrm{Sn}^{2+}} / \mathrm{Sn}}=-0.76+0.47 \\ }
\mathrm{E_{\text {cell }}^0 \mathrm{Sn} / \mathrm{Sn}^{2+}=29 \times 10^{-2} \\ }.
 

Posted by

Kshitij

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