\(E\, = \,1.51 - \,\frac{{0.059}}{5}\log \frac{{[M{n^{2 + }}]}}{{[MnO_4^ - ]\,\,{{[{H^ + }]}^8}}}\)
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 \(MnO_4^-\) \(/Mn^{2+}\).
$C{u^{2 + }}_{({C_1}aq)} + Zn(s) \Rightarrow Z{n^{2 + }}_{({C_2}aq)} + Cu(s)$ તાપમાને મુક્તઊર્જા ફેરફાર $\Delta G$ એ .... નું વિધેય છે.
$Zn ^{2+}+2 e ^{-} \rightarrow Zn ; E ^{\circ}=-0.760 \,V$
$Ag _{2} O + H _{2} O +2 e ^{-} \rightarrow 2 Ag +2 OH ^{-} ; E ^{\circ}=0.344 \,V$
જો $F$ $96,500 C mol ^{-1}$ હોય, તો કોષનો $\Delta G ^{\circ}$ શોધો. ($kJ mol ^{-1}$ માં)