- ✓$SO_2$
- B$SO_3$
- C$CO_2$
- D$NH_3$
Anode $\left\{\begin{array}{c}2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \\ 2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-}\end{array}\right\}$
Cathode $\left\{2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}-2 \mathrm{e}^{-}\right\} \times 3$
Net: $2 \mathrm{H}_{2} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}+\mathrm{O}_{2}+3 \mathrm{H}_{2}+6 \mathrm{H}^{+}+6 \mathrm{OH}^{-}$
Hence ratio of $\mathrm{n}_{\mathrm{O}_{2}}$ and $\mathrm{n}_{\mathrm{H}_{2}}$ is $1: 3 .$
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$Mn{O_2}(s)\, + \,4{H^ + }(aq)\, + \,2{e^ - }\, \to \, M{n^{ + 2}}(aq)\, + \,2{H_2}O(l)\,;$ $E_2^o\, = \,1.21\,\,V$
$MnO_4^{ - 1}(aq)\, + \,4{H^ + }(aq)\, + \,3{e^ - }\, \to \, Mn{O_2}(s)\, + \,3{H_2}O(l)\,;$ $E_3^o\,?$
value of $E_3^o$ will be ............ $\mathrm{V}$


$C{H_3}CHO\mathop {\xrightarrow{{(i)\,NaCN/HCl}}}\limits_{(ii)\,{H_2}O/{H^ \oplus }/\Delta } (A)\mathop {\xrightarrow{{Fenton}}}\limits_{reagent} (B)$
$(B)$ will be :