- ALead dioxide dissolves
- ✓Sulphuric acid is regenerated
- CThe lead electrode becomes coated with lead sulphate
- DThe amount of sulphuric acid decreases
Cathode $-\mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{PbO}_{2}+4 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}+2 \mathrm{e}^{-}$
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$(I)$ $BO_3^{-3} > CO_3^{-2} > NO_3^-$
$(II)$ $ClO_4^-> SO_4^{-2} > PO_4^{-3}$
$(III)$ $BF_3 > BCl_3 > BBr_3$
$(IV)$ $AlCl_3 > BCl_3$

$(a)\;\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}(\mathrm{OH}) \mathrm{CH}_{3} \stackrel{\mathrm{conc.H}, \mathrm{SO}_{4}}{\longrightarrow}$
$(b)\;\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \stackrel{\text { alc.KOH }}{\longrightarrow}$
$(c)\;\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \xrightarrow[\text { It should be }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CO}^{-} \mathrm{K}^{+}]{\text { given by } \mathrm{NTA}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{O}^{-} \mathrm{K}^{+}}$
$(d)\;\begin{array}{*{20}{c}}
{{{(C{H_3})}_2}C - C{H_2} - CHO} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{\Delta }$
Which of these reaction(s) will not produce Saytzeff product ?