At anode $2 \mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-}$
$E_{A g}=\frac{108}{1}$ ; $E_{O_{2}}=\frac{\frac{1}{2} \times 32}{2}=8$
$\frac{{{\text{W}}_{\text{Ag}}}}{{{\text{E}}_{\text{Ag}}}}=\frac{{{\text{W}}_{{{\text{O}}_{2}}}}}{{{\text{E}}_{{{\text{O}}_{2}}}}}\Rightarrow $ ${{\text{W}}_{\text{Ag}}}=\frac{1.6\times 108}{8}=21.6\,\text{gm}$
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$CN^+, CN^-, NO$ and $CN$
Which one of these will have the highest bond order?
$Pt ( s )\left| H _{2}( g , 1 bar )\right| HCl ( aq \cdot, pH =1)| AgCl ( s )| Ag ( s )$
The $pH$ of aq. HCl required to stop the photoelectric current from $K \left( w _{0}=2.25 eV \right),$ all other conditions remaining the same, is..........$\times 10^{-2}$ (to the nearest integer).
Given, $2.303 \frac{ RT }{ F }=0.06 V ; E _{ AgC1|Ag|C ^{-}}^{0}=0.22\, V$