\(1 atm\) \(\quad\) (at \(pH\; 10\))
If \(\mathrm{pH}=10\)
\(\mathrm{H}^{+}=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10}\)
From nernst equation, \(\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{0}-\frac{0.0591}{2} \log \frac{\left[\mathrm{H}^{+}\right]^{2}}{\mathrm{p}_{\mathrm{H}_{2}}}\)
For hydrogen electrode, \(E_{\text {cell }}^{0}=0\)
\(E_{\text {cell }}=-\frac{0.0591}{2} \log \frac{\left(10^{-10}\right)^{2}}{1}\)
\(0.0591 \times \log 10^{10}\)
\(0.0591 \times 10=0.591\; V\)
$C{u^{2 + }}_{({C_1}aq)} + Zn(s) \Rightarrow Z{n^{2 + }}_{({C_2}aq)} + Cu(s)$ તાપમાને મુક્તઊર્જા ફેરફાર $\Delta G$ એ .... નું વિધેય છે.
(આપેલ : $E _{ Zn ^{2+} \mid Zn }^{ o }=-0.763 V , E _{ Sn ^{x+} \mid Sn }^{ O }=+0.008 V$ ધારી લો $\frac{2.303 RT }{ F }=0.06\, V$ )
${Cu}_{({s})}+2 {Ag}^{+}\left(1 \times 10^{-3} \,{M}\right) \rightarrow {Cu}^{2+}(0.250\, {M})+2 {Ag}_{({s})}$
${E}_{{Cell}}^{\ominus}=2.97\, {~V}$
ઉપરની પ્રક્રિયા માટે ${E}_{\text {cell }}$ $=....\,V.$ (નજીકના પૂર્ણાંકમાં)
[આપેલ છે: $\log 2.5=0.3979, T=298\, {~K}]$