- A$\mathrm{CoCl}_{3}$
- B$\mathrm{FeCl}_{2}$
- C$\mathrm{CoCl}_{2}$
- ✓$\mathrm{FeCl}_{3}$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$Prussian blue
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$Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \rightleftharpoons $$2Ag_{(s)} + Zn^{2+}_{(aq)}+ 2OH^-_{(aq)}$
If half cell potentials are
$Zn^{2+}_{(aq)} + 2e^- \rightarrow Zn_{(s)}\,;\,\, E^o = - 0.76\, V$
$Ag_2O_{(s)} + H_2O_{(l)} + 2e^- \rightarrow 2Ag_{(s)} + 2OH^-_{(aq)}\,,$$E^o = 0.34\, V$
The cell potential will be ........... $V.$
($1$) $W$ and $\mathbf{X}$ are, respectively
$[A]$ $O_3$ and $P_4 O_6$ $[B]$ $O_2$ and $P_4 O_6$ $[C]$ $O_2$ and $P_4 O_{10}$ $[D]$ $O_3$ and $P_4 O_{10}$
($2$) $Y$ and $Z$ are, respectively
$[A]$ $N_2 O_3$ and $\mathrm{H}_3 \mathrm{PO}_4$
$[B]$ $N_2 O_5$ and $\mathrm{HPO}_3$
$[C]$ $N_2 O_4$ and $HPO_3$
$[D]$ $N_2 O_4$ and $H_3 PO_3$
Give the answer of quetion ($1$) and ($2$)