\(2{e^ - }\, + \,2{H_2}O\, \to \,{H_2}\, + \,2O{H^ - }\)
(valence factor) \(H_2=2\)
At \(NTP\) \(22400\,mL\) of \(H_2=1\,mole\) of \(H_2\,\,112\,mL\) of
\({H_2}\, = \,\frac{1}{{22400}}\, \times \,112\, = \,0.005\,mole\) of \(H_2\)
Moles of \(H_2\) produced
\( = \frac{{I\, \times t}}{{96500}}\, \times \) mole ratio
\(0.05 = \frac{{I\, \times 965}}{{96500}}\, \times \frac{{1\,\,mole\,\,of\,\,{H_2}}}{{2\,\,mole\,\,of\,\,{e^ - }}}\)
\(I\,=\,1.0\,A\)
$mol^{-1}, ᴧ^{0}\, KCl = 150\, S\, cm^{2}\, mol^{-1}$ હોય, તો $ᴧ^{0}\, NaBr$ .............. ${\rm{S}}\,{\rm{c}}{{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$ શોધો.
${Cu}_{({s})}+2 {Ag}^{+}\left(1 \times 10^{-3} \,{M}\right) \rightarrow {Cu}^{2+}(0.250\, {M})+2 {Ag}_{({s})}$
${E}_{{Cell}}^{\ominus}=2.97\, {~V}$
ઉપરની પ્રક્રિયા માટે ${E}_{\text {cell }}$ $=....\,V.$ (નજીકના પૂર્ણાંકમાં)
[આપેલ છે: $\log 2.5=0.3979, T=298\, {~K}]$