In a cell that utilises the reaction \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\) addition of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to cathode compartment, will

(C) increase the E and shift equilibrium to the right

(A) lower the E and shift equilibrium to the left

(B) increases the E and shift equilibrium to the left

(D) Lower the E and shift equilibrium to the right

JEE Main · 2004 · 24 Apr 2004 · Offline · Redox Reactions and Electrochemistry

In a cell that utilises the reaction

Zn (s) + 2 H^{+} (aq) ⟶ Zn^{2 +} (aq) + H_{2} (g)

addition of

H_{2} SO_{4}

to cathode compartment, will

Answer: (C)

Adding sulphuric acid to the cathode compartment raises the

H X^{+}

concentration, and

H X^{+}

is a reactant in

Zn (s) + 2 H X^{+} Zn X^{2 +} + H X_{2}

. By the Nernst equation a higher reactant concentration lifts the cell potential

E

, and by Le Chatelier's principle the extra

H X^{+}

pushes the equilibrium forward to the right. So

E

increases and the reaction shifts toward products.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option