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In a cell that utilises the reaction
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})$
addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ to cathode compartment, will
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$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})$
addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ to cathode compartment, will
Solution:
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Verified Answer
The correct answer is:
increase the $\mathrm{E}$ and shift equilibrium to the right.
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}+(\mathrm{aq}) \rightleftharpoons$
$$
\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})
$$
$$
\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\circ} \mathrm{cell}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{H}^{+}\right]^{2}}
$$
Addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ will increase $\left[\mathrm{H}^{+}\right]$ and $\mathrm{E}_{\text {cell }}$ will also increase and the equilibrium will shift towards RHS.
$$
\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})
$$
$$
\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\circ} \mathrm{cell}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{H}^{+}\right]^{2}}
$$
Addition of $\mathrm{H}_{2} \mathrm{SO}_{4}$ will increase $\left[\mathrm{H}^{+}\right]$ and $\mathrm{E}_{\text {cell }}$ will also increase and the equilibrium will shift towards RHS.
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