Question 5 of 6: Nickel electrodeposition: safe pH window and hydrogen co-evolution

Problem 5 — Nickel electrodeposition: safe pH window and hydrogen co-evolution (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the pH window that avoids both H2 evolution and Ni(OH)2 precipitation, across the full 0.01-1 M range; (b) a labelled electrolytic-cell circuit sketch; (c) the volume of H2(g) evolved per kg of Ni produced at pH 2, 70% current efficiency.

Approach. (a) Two independent Nernst/solubility inequalities bound pH from below (H2 evolution) and above (hydroxide precipitation); because $[\text{Ni}^{2+}]$ fluctuates, the SAFE window must satisfy both bounds at BOTH concentration extremes simultaneously. (b) A standard two-electrode electrolytic cell with a DC source. (c) Faraday's law applied twice — once to the desired Ni deposition (70% of charge) and once to the parasitic H2 evolution (30% of charge) that shares the same total charge.

1. (a-i) Lower pH bound — avoid H2 evolution. H2 evolution is avoided as long as the Ni deposition potential stays ABOVE (less negative than) the H+/H2 potential, $E_{\text{H}^+/\text{H}_2}=-0.0592\,\text{pH}$. Setting $E_{\text{Ni}^{2+}/\text{Ni}}=E_{\text{H}^+/\text{H}_2}$ at the boundary: $$-0.25+\dfrac{0.0592}{2}\log[\text{Ni}^{2+}]=-0.0592\,\text{pH}_{\min}\ \Rightarrow\ \text{pH}_{\min}=\dfrac{0.25-0.0296\log[\text{Ni}^{2+}]}{0.0592}.$$ At $[\text{Ni}^{2+}]=1$ M: $\text{pH}_{\min}=4.22$. At $[\text{Ni}^{2+}]=0.01$ M (more dilute, more negative $E_\text{Ni}$, so a HIGHER pH is needed to keep H2 suppressed): $\text{pH}_{\min}=5.22$. The worst case across the fluctuating range is the higher value, $\boxed{\text{pH}>5.22}$.
2. (a-ii) Upper pH bound — avoid Ni(OH)2 precipitation. From $K=[\text{H}^+]^2/[\text{Ni}^{2+}]$ (Ni(OH)2(s) and H2O both at unit/fixed activity), precipitation sets in once $[\text{H}^+]$ falls to $\sqrt{K\,[\text{Ni}^{2+}]}$, i.e. $$\text{pH}_{\max}=-\tfrac12\big(\log K+\log[\text{Ni}^{2+}]\big)=-\tfrac12(-12.73+\log[\text{Ni}^{2+}]).$$ At $[\text{Ni}^{2+}]=1$ M: $\text{pH}_{\max}=6.365$. At $[\text{Ni}^{2+}]=0.01$ M: $\text{pH}_{\max}=7.365$. The worst case (most restrictive, i.e. the concentrated 1 M feed) is the LOWER value, $\boxed{\text{pH}<6.37}$.
3. (a-iii) Combined safe window. Both bounds must hold at every concentration the feed can reach, so the operating window is the intersection: $$\boxed{5.22<\text{pH}<6.37}.$$ This is a narrow but workable band, entirely set by the dilute end of the range for the lower bound and the concentrated end for the upper bound.

(b) As sketched in Fig. 4: the cathode is wired to the NEGATIVE terminal of the DC power supply (electrons flow in, reducing Ni2+ to Ni metal), and the anode is wired to the POSITIVE terminal (electrons are pulled out, oxidizing water to O2 and H+). This polarity — cathode negative, anode positive — is specific to an ELECTROLYTIC cell (Problem 2e); a spontaneous galvanic cell would have the opposite terminal assignment.

4. (c) H2 evolved per kg of Ni. Charge to deposit 1 kg Ni (Faraday's law, $n=2$, $M_{\text{Ni}}=58.69$ g/mol): $$n_{\text{Ni}}=\dfrac{1000}{58.69}=17.04\ \text{mol},\qquad Q_{\text{Ni}}=n_{\text{Ni}}\times2F=17.04\times2\times96{,}485=3.288\times10^{6}\ \text{C}.$$ This is 70% of the TOTAL charge passed (since 30% goes to H2): $$Q_{\text{total}}=\dfrac{Q_{\text{Ni}}}{0.70}=4.697\times10^{6}\ \text{C},\qquad Q_{\text{H}_2}=0.30\,Q_{\text{total}}=1.409\times10^{6}\ \text{C}.$$ Moles of H2 ($n=2$ electrons per H2 molecule): $$n_{\text{H}_2}=\dfrac{Q_{\text{H}_2}}{2F}=\dfrac{1.409\times10^{6}}{2\times96{,}485}=7.30\ \text{mol}.$$ At STP (0°C, 1 atm — check assumption, since the exam does not state T/P): $$V_{\text{H}_2}=7.30\times22.4=\boxed{163.6\ \text{L H}_2\ \text{per kg Ni produced}}.$$