Question #44754

An electrochemical cell consists of $1.0 \text{ L}$ of an aqueous solution of a weak monoprotic acid \ce{HA} ($K_a = 1.0 \times 10^{-4}$) at $298 \text{ K}$. The initial concentration of \ce{HA} is $0.09 \text{ M}$. The cell is equipped with a Silver (\ce{Ag}) anode and a Platinum (\ce{Pt}) cathode. A constant effective DC voltage of $25 \text{ V}$ is applied across the electrodes.

During electrolysis, the following reactions occur: Anode: \ce{Ag(s) + A^{-}(aq) -> AgA(s) + e^{-}} Cathode: \ce{H^{+}(aq) + e^{-} -> 1/2 H2(g)}

The precipitated \ce{AgA(s)} does not interfere with the cell's electrical resistance. The cell constant $l/A$ is exactly $\frac{1}{9.65} \text{ cm}^{-1}$. Given the limiting molar conductivities: $\lambda^\infty(\ce{H^+}) = 350 \text{ S cm}^2 \text{ mol}^{-1}$ and $\lambda^\infty(\ce{A^-}) = 50 \text{ S cm}^2 \text{ mol}^{-1}$.

Assuming the cell operates purely under ohmic control (negligible back EMF and overpotential) and the degree of dissociation $\alpha \ll 1$, calculate the time required to collect $560 \text{ mL}$ of \ce{H2(g)} at STP ($273 \text{ K}, 1 \text{ atm}$).

(Given: $1 \text{ F} = 96500 \text{ C mol}^{-1}$, Molar volume of ideal gas at STP = $22.4 \text{ L mol}^{-1}$)