Explain the Brønsted-Lowry concept of acids, calculate the pH of a 0.001 M H2SO4 solution, define heterogeneous catalysts, and identify the strongest oxidizing and reducing agents based on standard el

Sulfuric acid is a strong acid and dissociates completely in water.

• The strongest oxidizing agent is the species on the left side of the half-reaction with the most positive (or least negative) $E^\circ$ value. This species is most easily reduced.
• The strongest reducing agent is the species on the right side of the half-reaction with the most negative (or least positive) $E^\circ$ value. This species is most easily oxidized.

Given electrode potentials:
A: $Pb^{2+}(aq) + 2e^- \to Pb(s)$, $E^\circ = -0.13 \text{ V}$
B: $I_2(aq) + 2e^- \to 2I^-(aq)$, $E^\circ = +0.54 \text{ V}$
C: $MnO_4^-(aq) + 8H^+(aq) + 5e^- \to Mn^{2+}(aq) + 4H_2O(l)$, $E^\circ = +1.51 \text{ V}$

Comparing the $E^\circ$ values: $+1.51 \text{ V}$ is the most positive.
The strongest oxidizing agent is $MnO_4^-(aq)$.
Comparing the $E^\circ$ values: $-0.13 \text{ V}$ is the most negative.
The strongest reducing agent is $Pb(s)$.

Strongest oxidizing agent: $\boxed{MnO_4^-(aq)}$
Strongest reducing agent: $\boxed{Pb(s)}$

For a spontaneous reaction, the species with the more positive $E^\circ$ will be reduced (cathode), and the species with the more negative $E^\circ$ will be oxidized (anode).
Cell A: $Pb^{2+}(aq) + 2e^- \to Pb(s)$, $E^\circ = -0.13 \text{ V}$
Cell B: $I_2(aq) + 2e^- \to 2I^-(aq)$, $E^\circ = +0.54 \text{ V}$

Since $E^\circ_B (+0.54 \text{ V})$ is more positive than $E^\circ_A (-0.13 \text{ V})$, reaction B will occur at the cathode (reduction) and reaction A will occur at the anode (oxidation, reversed).

Anode (oxidation): $Pb(s) \to Pb^{2+}(aq) + 2e^-$
Cathode (reduction): $I_2(aq) + 2e^- \to 2I^-(aq)$

In this cell, $Pb(s)$ is oxidized at the anode, and $I_2(aq)$ is reduced at the cathode.
Therefore, electrons will flow from the lead electrode (Pb) to the iodine electrode (I$_2$).