In an electrolysis cell similar to the one employed in this experiment, a
student observed that his unknown metal anode lost \(0.208 \mathrm{g}\) while a
total volume of \(96.30 \mathrm{mL}\) of \(\mathrm{H}_{2}\) was being produced.
The temperature in the laboratory was \(25^{\circ} \mathrm{C}\) and the
barometric pressure was \(748 \mathrm{mm}\) Hg. At \(25^{\circ} \mathrm{C}\) the
vapor pressure of water is \(23.8 \mathrm{mm}\) Hg. To find the equivalent mass
of his metal, he filled in the blanks below. Fill in the blanks as he did.
$$P_{\mathrm{II}_{2}}=P_{\mathrm{har}}-V P_{\mathrm{H}_{2} \mathrm{O}}=$$
$$\operatorname{mm} \mathrm{Hg}=$$ atm $$V_{\mathrm{H}_{2}}=$$
$$\mathrm{mL}=$$ $$\mathbf{L}$$ $$T=$$ K$$n_{\mathrm{H}_{2}}=$$ moles
$$\left.n_{\mathrm{H}_{2}}=\frac{P V}{R T} \quad \text { (where }
P=P_{\mathrm{H}_{2}}\right)$$ 1 mole \(\mathrm{H}_{2}\) requires passage of
_faradays No. of faradays passed \(=\) Loss of mass of metal anode \(=\) g No.
grams of metal lost per faraday passed \(=\frac{\text { no. grams lost }}{\text
{ no. faradays passed }}=\quad \mathrm{g}=\mathrm{EM}\) The student was told
that his metal anodc was madc of iron. MM Fe = \(g\) The charge \(n\) on the Fe
ion is therefore $$P_{\mathrm{II}_{2}}=P_{\mathrm{har}}-V P_{\mathrm{H}_{2}
\mathrm{O}}=$$ $$\operatorname{mm} \mathrm{Hg}=$$ atm $$V_{\mathrm{H}_{2}}=$$
$$\mathrm{mL}=$$ $$\mathbf{L}$$ $$T=$$ K$$n_{\mathrm{H}_{2}}=$$ moles
$$\left.n_{\mathrm{H}_{2}}=\frac{P V}{R T} \quad \text { (where }
P=P_{\mathrm{H}_{2}}\right)$$
