Q. In a cell that utilizes the reaction $Zn(s) + 2H^{+}(aq)\ \rightarrow {Zn}^{2 +}(aq) + H_{2}(g)$ addition of $H_{2}{SO}_{4}$ to cathode compartment will
A) Lower the $E$ and shift equilibrium to the right
B) Lower the $E$ and shift equilibrium to the left
C) Increase the $E$ and shift equilibrium to the right
D) Increase the $E$ and shift equilibrium to the left
Explanation
(c)
$Zn(s) + 2H^{+}(aq) \rightleftharpoons {Zn}^{2 +}(aq) + H_{2}(g)$
$E_{cell} = E_{cell}^{{^\circ}} - \frac{0.059}{2}\log\frac{\left\lbrack {Zn}^{2 +} \right\rbrack}{\left\lbrack H^{+} \right\rbrack^{2}}$
When $H_{2}{SO}_{4}$ is added then $\left\lbrack H^{+} \right\rbrack$ will increase therefore $E_{cell}$ will also increase and equilibrium will shift towards right
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.