(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ કરો) $[$ આપેલ $\left.: \frac{2.303 RT }{ F }=0.059\right]$
\(MnO _{4}^{-}+ H ^{\oplus}+5 e ^{-} \rightarrow Mn ^{+2}+4 H _{2} O\)
Nernst equation:
\(E _{\text {cell }}= E _{ Cell }^{0}-\frac{0.059}{5} \log \frac{\left[ Mn ^{+2}\right]}{\left[ MnO _{4}^{-}\right]}\left[\frac{1}{ H ^{+}}\right]^{8}\)
\((I)\) Given \(\left[ H ^{\oplus}\right]=1\, M\)
\(E _{1}= E ^{0}-\frac{0.059}{5} \log \frac{\left[ Mn ^{+2}\right]}{\left[ MnO _{4}^{-}\right]}\)
\((II)\) Now : \(\left[ H ^{\oplus}\right]=10^{-4}\, M\)
\(E _{2}= E ^{0}-\frac{0.059}{5} \log \frac{\left[ Mn ^{+2}\right]}{\left[ MnO _{4}^{-}\right]} \times \frac{1}{\left(10^{-4}\right)^{8}}\)
\(= E ^{0}-\frac{0.059}{5} \log \frac{ Mn ^{+2}}{\left[ MnO _{4}^{-}\right]}+\frac{0.059}{5} \log 10^{-32}\)
therefore : \(\left| E _{1}- E _{2}\right|=\frac{0.059}{5} \times 32\)
\(=0.3776\, V =3776 \times 10^{-4}\)
\(x =3776\)
$Zn ^{2+}+2 e ^{-} \rightarrow Zn ; E ^{\circ}=-0.760 \,V$
$Ag _{2} O + H _{2} O +2 e ^{-} \rightarrow 2 Ag +2 OH ^{-} ; E ^{\circ}=0.344 \,V$
જો $F$ $96,500 C mol ^{-1}$ હોય, તો કોષનો $\Delta G ^{\circ}$ શોધો. ($kJ mol ^{-1}$ માં)