$Zn | Zn ^{2+}(0.1\, M ) \| Ag ^{+}(0.01 \,M )| Ag$
[આપેલ $: E_{Z n^{+2}/Z_{n}}^{0}=-0.76 \,V ; E _{A g^{+} / A_{ g }}^{0}=+0.80 \,V ; \frac{2.303 RT }{ F }=0.059$]
$2 Ag ^{+}_{( aq. )}+2 e ^{-} \rightarrow 2 Ag _{( s )}$
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$Zn _{( s )}+2 Ag ^{+}_{( aq. )} \rightarrow Zn _{( aq .)}^{2+}+ 2 Ag _{( s )}$
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$E _{ cell }^{0}= E _{ Ag ^{+} / Ag }^{0}- E _{ zn ^{2+} / Zn }^{0}$
$=0.80-(-0.76)$
$=1.56 \,V$
$E _{\text {cell }}=1.56 \frac{-0.059}{2} \log \frac{\left[ Zn ^{2+}\right]}{\left[ Ag ^{+}\right]^{2}}$
$=1.56-\frac{0.059}{2} \log \frac{0.1}{(0.01)^{2}}$
$=1.56-\frac{0.059}{2} \times 3$
$=1.56-0.0885$
$=1.4715$
$=147.15 \times 10^{-2}$
$Cr\, | \,Cr^{3+}_{(0.1\,M)}\,||\, Fe^{2+}_{(0.01\, M)}\,|\, Fe$
$P{b^4} + 2{e^ - } \longrightarrow P{b^{2 + }};\,{E^o} = + 1.67\,V$
$C{e^{4 + }} + {e^ - } \longrightarrow C{e^{3 + }};\,{E^o} = + 1.61\,V$
$B{i^{3 + }} + 3{e^ - } \longrightarrow Bi;\,{E^o} = + 0.20\,V$ આપેલ છે. તો આ ઘટકતી ઓક્સિડેશતકર્તા તરીકેની ક્ષમતા ક્યા ક્રમમાં વધશે?
$\Delta G_{f}^{o}\left(A g_{2} O\right)=-11.21\, kJ\,mol ^{-1}$
$\Delta G_{f}^{o}(Z n O)=-318.3\, kJ \,mol ^{-1}$
ત્યારે $E^{o}$કોષ નો બટન શેલ.........$V$ શું હશે ?
($Cu$નું આણ્વિય દળ $63.5\, amu$)