
- A$K_1 = K_2$
- ✓$K_1 > K_2$
- C$K_2 > K_1$
- D$K_1 = K_2 = 1$


Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${[Fe\,{(CN)_6}]^{4 - }}\, \to \,{[Fe{(CN)_6}]^{3 - }}\, + \,{e^ - }\,;\,$ ${E^o}\, = \, - \,0.35\,V$
$\,F{e^{2 + }}\, \to \,F{e^{3 + }}\, + \,{e^ - }\,;$ ${E^o}\, = \, - \,0.77\,V$

| $Pb ^{2+} / Pb$ | $-0.13 V$ |
| $Ni ^{2+} / Ni$ | $-0.24 V$ |
| $Cd ^{2+} / Cd$ | $-0.40 V$ |
| $Fe ^{2+} / Fe$ | $V-0.44 V$ |
To a solution containing $0.001 M$ of $X ^{2+}$ and $0.1 M$ of $Y ^{2+}$, the metal rods $X$ and $Y$ are inserted (at $298 K$ ) and connected by a conducting wire. This resulted in dissolution of $X$. The correct combination(s) of $X$ and $Y$, respectively, is (are)
(Given: Gas constant, $R =8.314 J K ^{-1} mol ^{-1}$,
Faraday constant, $F =96500 C mol ^{-1}$ )
$(A)$ $Cd$ and $Ni$ $(B)$ $Cd$ and $Fe$ $(C)$ $Ni$ and $Pb$ $(D)$ $Ni$ and $Fe$

If one Faraday of electricity is passed through the battery during the charging, the number of moles of $Cr^{3+}$ removed from the solution is
