\((152\;sc{m^2}) \wedge _{KBr}^0\; = \; \wedge _{{K^ + }}^0 + \wedge _{B{r^ - }}^0\) ..... \((2)\)
\((150\;sc{m^2}) \wedge _{KCl}^0\; = \; \wedge _{{K^ + }}^0 + \wedge _{C{l^ - }}^0\) ..... \((3)\)
By equation \((1)+(2) \,-\,(3)\)
\( \wedge _{NaBr}^0\; = \; \wedge _{N{a^ + }}^0 + \wedge _{B{r^ - }}^0\)
\(= 126 + 152 - 150 = 128\) \(Scm^2 mol^{-1}\)
$Mn^{2+} +2e- \rightarrow Mn;\, E^o = -1.18\,V$
$2(Mn^{3+} +e^- \rightarrow Mn^{2+} )\,;\,E^o=+1.51\,V$
તો $3Mn^{2+} \rightarrow Mn+ 2Mn^{3+}$ માટે $E^o$ કેટલો થશે ?
$A$. $\mathrm{Fe}$ $B$. $\mathrm{Mn}$ $C$. $\mathrm{Ni}$ $D$. $\mathrm{Cr}$ $E$. $\mathrm{Cd}$
Choose the correct answer from the options given below:
$(A)$ $Sn^{+4}+ 2e^{-} \rightarrow Sn^{2+}$, $E^o= + 0.15\,V$
$(B)$ $2Hg^{+2} + 2e^{-} \rightarrow Hg_{2}^{+2}$, $E^o = + 0.92\,V$
$(C)$ $PbO_2 + 4H^{+} + 2e^{-} \rightarrow Pb^{+2} + 2H_2O$, $E^o = + 1.45\,V$
$Pt ( s )\left| H _2( s )( latm )\right| H ^{+}\left( aq ,\left[ H ^{+}\right]=1\right)|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid \operatorname{Pt}( s )$
આપેલ : $E _{ Fe ^{3+} / e ^{2 *}}^0=0.771\,V$ અને $E _{ H ^{+}+\frac{1}{2} H _2}^0=0\,V , T =298\,K$
જો કોષનો પોટેન્શિયલ $0.712\,V$, હોય, તો $Fe ^{-2}$ થી $Fe ^{+3}$ ની સાંદ્રતાની ગુણોત્તર છે.(નજીકનો પૂર્ણાંક)