- A$SnO$
- B$PbO$
- C$PbO_2$
- ✓$N_2O$
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$\text{H}^+(\text{aq})+\text{e}^-\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{1}{2}\text{H}_2(\text{g});\ \ \ \ \ \text{E}^\ominus_{\text{Cell}}=0.00\text{V}$
$2\text{H}_2\text{O}_{(\text{l})}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{O}_{2(\text{g})}+4\text{H}^{+}_{(\text{aq})}+4\text{e}^-;\ \ \ \ \ \text{E}^0_{\text{Cell}}=1.23\text{V}$
$2\text{SO}^{2-}_{4(\text{aq})}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{S}_2\text{O}^{2-}_{8(\text{aq})}+2\text{e}^-;\ \ \ \ \ \text{E}^0_{\text{cell}}=1.96\text{V}$
(Given: Atomic number: $V , 23 ; Cr , 24 ; Fe , 26;Ni , 28)$
$V ^{3+} \cdot Cr ^{3+}, Fe ^{2+}, Ni ^{3+}$

${C_2}{H_5}Br\xrightarrow{X}{\text{product}}\xrightarrow{{\text{Y}}}{C_3}{H_7}N{H_2}$
$B + {H_2}O \to C + HCl$
Compound $(A),\,(B)$ and $(C)$ will be respectively