$E\, = \,1.51 - \,\frac{{0.059}}{5}\log \frac{{[M{n^{2 + }}]}}{{[MnO_4^ - ]\,\,{{[{H^ + }]}^8}}}$
Taking $Mn^{2+}$ and $MnO_4^-$ in standard state i.e.$1\,M,$
$E\, = \,1.51 - \,\frac{{0.059}}{5}\, \times \,\,8\,\,\log \frac{1}{{\,[{H^ + }]}}$$\, = \,1.51 - \,\frac{{0.059}}{5}\, \times \,\,8\, \times \,3 = 1.2268\,\,V$
Hence at this $pH,$ $MnO_4^-$ will oxidise only $Br^-$ and $I^-$ as $SRP$ of $Cl_2/Cl^-$ is $1.36\,V$ which is greater than that for $MnO_4^-$ $/Mn^{2+}$.
$298\,K$ પર જ્યારે $\frac{\left[M^*(a q)\right]}{\left[M^{3 *}(a q)\right]}=10^a$ હોય ત્યારે આપેલ કોષ નો $E_{\text {cell }}$ એ $0.1115\,V$ છે. $a$ નું મૂલ્ય $............$ છે.આપેલ : $E _{ M }^\theta{ }^{3+} M ^{+}=0.2\,V$
$\frac{2.303\,R T}{F}=0.059\,V$