- ✓$SO_2$ is evolved
- BLead sulphate is produced at both electrodes
- CSulphuric acid is consumed
- DWater is formed
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$(I)$ $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}$
$(II)$ $ \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
$(III)$ $\mathrm{Na}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\left(\Delta_{0}>\mathrm{P}\right)$
$(IV)$ $\left(\mathrm{Et}_{4} \mathrm{N}\right)_{2}\left[\mathrm{CoCl}_{4}\right]$
$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.02 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(0.04 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$
$\mathrm{E}_{\text {cell }}=...... \,\times 10^{-2} \,\mathrm{~V} { (Nearest integer) }$
${\left[\text { Use }: \mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{0}=-0.34\, \mathrm{~V}, \mathrm{E}_{2 \mathrm{n} / \mathrm{Zn}^{2+}}^{0}=+0.76 \,\mathrm{~V}\right.}$
$\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059\, \mathrm{~V}\right]$
Phenol $\xrightarrow[{{H^ + }}]{{{\text{CHC}}{{\text{l}}_3}{\text{ /NaOH}}}}$ Salicyladehyde
$2NO + 2H_2 \longrightarrow N_2 + 2H_2O$
has following mechanism
Step $-I$ : $2NO \longrightarrow N_2O_2$
Step $-II$ : $N_2O_2 + H_2 \longrightarrow N_2O + H_2O$
Step $-III$ : $N_2O + H_2 \longrightarrow N_2 + H_2O$
Which of the following substance is a reaction intermediate
