${E_{cell}}\, = \,\,E_{cell}^ \circ \, + \,\,\frac{{0.059}}{2}\log \,\,\frac{{[F{e^{ + 2}}]}}{{[Z{n^{ + 2}}]}}$
$0.2905\,\, = \,\,E_{cell}^ \circ \, + \,\,\frac{{0.059}}{2}\,\log \,\frac{{0.01}}{{0.1}}$
$\therefore \,E_{cell}^ \circ \, = \,\,0.2905\,\, + \,\,0.0295\,\, = \,\,0.32\,\,volt$
$Now,\,\,\,E_{cell}^ \circ \, = \,\,\frac{{0.059}}{2}\,\,\log \,\,{K_c}$
$0.32\,\, = \,\,\frac{{0.059}}{2}\,\log \,\,{K_c}$
$KC\,\, = \,\,{10^{0.32/0.0295}}$
$AgI$ માટે $log\, K_{sp}$ નું મૂલ્ય શું હશે? (જ્યાં $K_{sp}=$ દ્રાવ્યતા નીપજ)
(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ કરો) $[$ આપેલ $\left.: \frac{2.303 RT }{ F }=0.059\right]$
$I$. $\log \,\,K\, = \,\frac{{nF{E^o}}}{{2.303\,RT}}$
$II$. $K\, = \,{e^{\frac{{nF{E^o}}}{{RT}}}}$
$III$. $\log \,\,K\, = -\,\frac{{nF{E^o}}}{{2.303\,RT}}$
$IV$. $\log \,\,K\, = 0.4342\,\,\frac{{-nF{E^o}}}{{RT}}$
સાચું વિધાન $(s)$ પસંદ કરો