Zinc reacts with dil. $H_{2} S O_{4}$ to give $H_{2}$ . It also reacts with conc. $H_{2} S O_{4}$ to give $S O_{2}$ .
If true enter 1, else enter 0.

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Solution

$(a) Z n + H_{2} S O_{4} d i l \to Z n S O_{4} + H_{2} In this reaction, H^{+} is reduced by Z n . Z n + 2 H^{+} \to H_{2} + Z n^{2 +} (b) Z n + 2 H_{2} S O_{4} c o n c . \to Z n S O_{4} + S O_{2} + 2 H_{2} O Z n r e d u c e s S O_{4}^{2 -} t o S O_{2} . S O_{4}^{2 -} + 4 H^{+} + Z n \to Z n^{2 +} + S O_{2} + 2 H_{2} O$
Hence, the given statement is true.

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