Which statement (① to ⑥) about batteries and electrolysis is correct?
③ is correct. By Faraday’s law, for a fixed amount of electricity $Q$, the amount of substance produced/consumed is $n=\frac{Q}{zF}$, so the moles changed are inversely proportional to the ion’s valence $z$, independent of the ion’s identity.
What you are checking in these options
Each statement is about where oxidation or reduction happens, how pH changes near electrodes, or how Faraday’s law and electrode potentials control what forms during electrolysis.
Testing each option quickly
① Lead storage battery during charging: the negative electrode of the battery is the cathode during charging, so reduction occurs there, not oxidation. So ① is false.
② Electrolysis of dilute $H_2SO_4$ with Pt electrodes: at the anode, water is oxidized: $$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$ $H^+$ increases near the anode, so pH decreases, not increases. So ② is false.
③ Same charge passed through an electrolyte: Faraday’s law gives $$n = \frac{Q}{zF}$$ So for the same $Q$, the moles that react are proportional to $1/z$ regardless of the ion type. So ③ is true.
④ Hydrogen-oxygen fuel cell (phosphoric acid electrolyte): the negative electrode (anode) uses hydrogen as the active material; oxygen is at the positive electrode (cathode). So ④ is false.
⑤ Ion-exchange membrane method for $NaOH$ (chlor-alkali): the membrane allows mainly $Na^+$ to pass from anode side to cathode side; $OH^-$ does not migrate through it as the main transported ion. So ⑤ is false.
⑥ Molten mixture of $MgCl_2$ and $NaCl$: the easier-to-reduce cation is $Mg^{2+}$ because $E^\circ(Mg^{2+}/Mg)\approx -2.37\, V$ is more positive than $E^\circ(Na^+/Na)\approx -2.71\, V$. So magnesium forms before sodium, making ⑥ false.
Final selection
Only ③ is correct.
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