$\Delta G_{f}^{o}\left(A g_{2} O\right)=-11.21\, kJ\,mol ^{-1}$
$\Delta G_{f}^{o}(Z n O)=-318.3\, kJ \,mol ^{-1}$
ત્યારે $E^{o}$કોષ નો બટન શેલ.........$V$ શું હશે ?
\(\Delta G_{f}^{o}\left(A g_{2} O\right)=-11.21 kJmol ^{-1}\)
\(\Delta G_{f}^{o}(Z n O)=-318.3\, kJmol ^{-1}\)
As we know that,
\(\Delta G^{o}=\Delta G_{f}^{o}(Z n O)-\Delta G_{f}^{o}\left(A g_{2} O\right)\)
Putting the values in the above given equation, we get
\(\therefore \Delta G^{o}=(-318.30+11.21) \,kJ \,mol ^{-1}\)
\(=-307.09 \,kJ =-307.09 \times 10^{3} \,J\)
\(\because \Delta G^{o}=-n F E_{c e l l}^{o}\)
\(\therefore-307.09 \times 10^{3}=-2 \times 96500 \times E_{c e l l}^{o}\)
or \(E_{\text {cell}}^{o}=1.591\, V\)
${Cu}({s})\left|{Cu}^{2+}({aq})(0.01 {M}) \| {Ag}^{+}({aq})(0.001 {M})\right| {Ag}({s})$ કોષ માટે ,કોષનો પોટેન્શિયલ $=.....\times 10^{-2} {~V}$
[ઉપયોગ : $\frac{2.303 {RT}}{{F}}=0.059$ ]
$\lambda_{\mathrm{m}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{x} \;\mathrm{S}\; \mathrm{cm}^{2} \mathrm{mol}^{-1}$
$\lambda_{\mathrm{m}\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{y} \;\mathrm{S\;cm}^{2} \mathrm{mol}^{-1}$
$\lambda_{\mathrm{m}(\mathrm{CH_3} \mathrm{COOK})}^{0}=\mathrm{z}\; \mathrm{S\;cm}^{2} \mathrm{mol}^{-1}$
$\mathrm{CH}_{3} \mathrm{COOH}$ માટે $\lambda_{\mathrm{m}}^{0}\left(\mathrm{in}\; \mathrm{S} \;\mathrm{cm}^{2} \mathrm{mol}^{-1}\right)$ શું હશે ?
(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ કરો) $[$ આપેલ $\left.: \frac{2.303 RT }{ F }=0.059\right]$