MCQ
Perdisulphuric acid has the following bond
- ✓$O\, \leftarrow \,O\,\, = \,\,O$
- B$ \leftarrow \,O\,\, = \,\,O \rightarrow $
- C$ > \,O \to \,O\, < $
- D$ - \,\,O\, - \,O\, - $
$H _2 O > H _2 S > H _2 Se > H _2 Te$
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$\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Pb}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$
the ratio $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$ when this cell attains equilibrium is
(Given $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0.14 \mathrm{\;V}$ $\left.\mathrm{E}_{\mathrm{Pb}^{+2}/{\mathrm{Pb}}}^{0}=-0.13 \;\mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)$


Reason : Lower the activation energy, faster is the reaction.