Which species is the weakest oxidizing agent, given the following standard electrode potentials: R2+/R = -2.38 V, S+/S2+ = +2.87 V, T2+/T = +0.34 V, U2+/U = -2.90 V, V+/V2+ = +0.80 V?

An oxidizing agent is a species that gains electrons (gets reduced) in a redox reaction. The standard electrode potential ($E^\circ$) measures the tendency of a species to be reduced.

A more positive $E^\circ$ indicates a stronger tendency for reduction, meaning the species is a stronger oxidizing agent. Conversely, a more negative $E^\circ$ indicates a weaker tendency for reduction, meaning the species is a weaker oxidizing agent.

From the table, the standard electrode potentials for the oxidizing agents (species on the left side of the half-equations) are:

1. R$^{2+}$: $-2.38$ V
2. $\frac{1}{2}$S$^{+}$: $+2.87$ V
3. T$^{2+}$: $+0.34$ V
4. U$^{2+}$: $-2.90$ V
5. $\frac{1}{2}$V$^{+}$: $+0.80$ V

The most negative $E^\circ$ value is $-2.90$ V, which corresponds to U$^{2+}$.

The weakest oxidizing agent is U$^{2+}$.
Explanation: U$^{2+}$ has the most negative standard electrode potential ($E^\circ = -2.90$ V), indicating it has the least tendency to gain electrons and be reduced. Therefore, it is the weakest oxidizing agent among the given species.

II)

The electromotive force (e.m.f) of a cell is calculated as $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$. To produce the highest e.m.f, we need to select the half-cell with the most positive $E^\circ$ to act as the cathode (where reduction occurs) and the half-cell with the most negative $E^\circ$ to act as the anode (where oxidation occurs).

The most positive $E^\circ$ value in the table is $+2.87$ V, corresponding to the half-cell:

The most negative $E^\circ$ value in the table is $-2.90$ V, corresponding to the half-cell:

The two half-cells that will produce the highest e.m.f are:

• $\frac{1}{2}\text{S}^{+}(\text{aq}) + \text{e}^{-} \rightleftharpoons \text{S(s)}$
• $\text{U}^{2+}(\text{aq}) + 2\text{e}^{-} \rightleftharpoons \text{U(s)}$

The highest e.m.f produced would be: