(આપેલ: $ E^oCr^{+3}| Cr = -0.75 \,V$ $E^o Fe^{+2} | Fe = - 0.45\, V)$
એનોડ પર,\( \Rightarrow \,\,[Cr \to C{r^{ + 3}} + 3{e^ - }]\,\, \times \,\,2\)
કેથોડ પર \( \Rightarrow \,\,[F{e^{ + 2}} + 2{e^ - } \to Fe]\,\, \times \,\,3\)
સંપૂર્ણ પ્રક્રિયા \({2Cr + 3F{e^{ + 2}} \to 2C{r^{ + 3}} + 3Fe}\)
\(E_{cell}^ o= \) ઓક્સિડેશન પોટેન્શિયલ \(+\) રિડકશન પોટેન્શિયલ
\( = 0.75+( -\,0.45)=0.30\)
\({E_{cell}}\, =E_{cell}^o - \frac{{0.059}}{n}\,\log \) [પ્રક્રિયા]/[પ્રક્રિયા]
\( = 0.30- \frac{{0.059}}{6}\,\log \frac{{{{[C{r^{ + 3}}]}^2}}}{{{{[F{e^{ + 2}}]}^3}}}\)
\( = \,0.30\, - \frac{{0.059}}{6}\,\log \,\frac{{{{[0.1]}^2}}}{{{{[0.01]}^3}}}\)
\(\, = \,0.30\, - \,\frac{{0.24}}{6}\,= \,0.26\,Volt.\)
$mol^{-1}, ᴧ^{0}\, KCl = 150\, S\, cm^{2}\, mol^{-1}$ હોય, તો $ᴧ^{0}\, NaBr$ .............. ${\rm{S}}\,{\rm{c}}{{\rm{m}}^2}{\rm{mo}}{{\rm{l}}^{ - 1}}$ શોધો.
${[Fe\,{(CN)_6}]^{4 - }}\, \to \,{[Fe{(CN)_6}]^{3 - }}\, + \,{e^ - }\,;\,$ ${E^o}\, = \, - \,0.35\,V$
$\,F{e^{2 + }}\, \to \,F{e^{3 + }}\, + \,{e^ - }\,;$ ${E^o}\, = \, - \,0.77\,V$
$H_3PO_4 + OH^- \rightarrow H_2PO_4^- + H_2O$ ;
$H_3PO_4 + 2OH^- \rightarrow HPO_4^{2-} + 2H_2O$ ;
$H_3PO_4 + 3OH^- \rightarrow PO_4^{3-} + 3H_2O$