Question

In a cell that utilizes the reaction, what will happen if we add $H_{2}S{O}_4$ to cathode compartment?
${\left(\text{Zn}\right)}_{\left(\text{s}\right)}+2{\text{H}}_{\left(\text{aq}\right)}^{+}\to {\text{Zn}}_{\left(\text{aq}\right)}^{2+}+{\left(\text{H}\right)}_{2\left(\text{g}\right)}$

• A. Lower the E and shift equilibrium to the left
• B. Lower the E and shift the equilibrium to the right
• C. Increase the E and shift the equilibrium to the right
• D. Increase the E and shift the equilibrium to the left

Answer 1

Answer: (C)

${\left(\text{Zn}\right)}_{\left(\text{s}\right)}+2{\text{H}}_{\left(\text{aq}\right)}^{+}\rightleftharpoons {\text{Zn}}_{\left(\text{aq}\right)}^{2+}+{\left(\text{H}\right)}_{2\left(\text{g}\right)}$
${\text{E}}_{\text{ cell}}={\text{E}}_{\text{ cell}}^{\text{o}}-\frac{\text{0.059}}{2}\text{ log }\frac{\left[{\text{Zn}}^{2+}\right]\times {\text{p}}_{{\text{H}}_2}}{{\left[{\text{H}}^{+}\right]}^2}$
On adding ${\text{H}}_2{\text{SO}}_4$ the $\left[{\text{H}}^{+}\right]$ will increase therefore ${\text{E}}_{\text{ cell}}$ will also increase and the equilibrium will shift towards the right.