Dihydrogen of high purity (> 99.95%) is obtained through:

We have to choose that experimental procedure which delivers dihydrogen whose purity is higher than $$99.95\%$$. In other words, the method must minimise every possible gaseous or vapour impurity that could mix with the $$H_{2}$$ produced.

First we examine the chemical route given in option A. The reaction involved is

$$Zn + 2\,HCl(dil.) \;\longrightarrow\; ZnCl_{2} + H_{2}\uparrow$$

During this laboratory preparation, water vapour from the acid and traces of chloride or even sulphur-containing impurities commonly present in zinc can accompany the evolved hydrogen. These side components bring the purity well below $$99.95\%$$, so option A cannot satisfy the requirement.

Now let us analyse option B, namely the electrolysis of acidified water using platinum (Pt) electrodes. The pertinent half-reactions are

At the cathode: $$2\,H^+ + 2e^- \;\longrightarrow\; H_{2}(g)$$

At the anode: $$2\,H_{2}O \;\longrightarrow\; O_{2}(g) + 4\,H^+ + 4e^-$$

Although platinum is inert, small bubbles of oxygen generated at the anode can diffuse through the electrolyte and contaminate the hydrogen collected at the cathode. Acid mists may also be entrained. The gas produced is therefore not of the exceptionally high purity demanded.

We proceed to option C, the electrolysis of brine (aqueous $$NaCl$$). The half-cell processes are

At the cathode: $$2\,H_{2}O + 2e^- \;\longrightarrow\; H_{2}(g) + 2\,OH^-$$

At the anode: $$2\,Cl^- \;\longrightarrow\; Cl_{2}(g) + 2e^-$$

Chlorine gas is formed simultaneously with hydrogen and can mix with it. Even highly efficient cell designs cannot eliminate slight chlorine carry-over, so the hydrogen obtained from brine electrolysis does not attain $$99.95\%$$ purity.

Finally, we consider option D, the electrolysis of a warm, dilute $$Ba(OH)2$$ solution employing nickel (Ni) electrodes. The reactions are

At the cathode: $$2\,H_{2}O + 2e^- \;\longrightarrow\; H_{2}(g) + 2\,OH^-$$

At the anode: $$4\,OH^- \;\longrightarrow\; O_{2}(g) + 2\,H_{2}O + 4e^-$$

Crucially, barium hydroxide is a strong base whose cation $$Ba^{2+}$$ and anion $$OH^-$$ remain completely in the liquid phase; they are non-volatile and therefore cannot accompany the gaseous products. Nickel electrodes are sturdy yet practically inert under these conditions, so they introduce no foreign gases. The only potential contaminant is oxygen, but the solubility of oxygen in the basic medium is low and the cell can be designed so that the hydrogen outlet is physically isolated from the oxygen-evolution compartment. Consequently the hydrogen collected possesses a purity better than $$99.95\%$$, the standard demanded for applications such as the electronic and metallurgical industries.

Summarising the discussion, options A, B and C each introduce impurities (acid mist, oxygen, chlorine, etc.) that limit hydrogen purity, whereas option D yields dihydrogen of the required exceptionally high purity.

Hence, the correct answer is Option D.