Identify the weakest reducing agent and its standard reduction potential from the provided table. Also, identify the element likely to be hydrogen and explain why.

A reducing agent is a species that gets oxidized. The weakest reducing agent is the one that is most difficult to oxidize, meaning its standard reduction potential ($E^\circ$) is the most positive.

The most positive $E^\circ$ value is $+2.87$ V for the half-reaction $Q^{3+}(aq) + 3e^- \rightleftharpoons Q(s)$.
The weakest reducing agent is $Q(s)$. The question asks for the $E^\circ$ value associated with it.

The $E^\circ$ value of the weakest reducing agent is $\boxed{\text{+2.87 V}}$.

a) (ii)

The standard electrode potential for hydrogen ($2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$) is defined as 0.00 V.

The element with an $E^\circ$ value of 0.00 V is $T$.

Element $T$ is likely to be hydrogen.
Reason: The standard electrode potential of hydrogen is defined as 0.00 V, which matches the $E^\circ$ value for element $T$.

a) (iii)

For elements $S$ and $V$:
$S^{2+}(aq) + 2e^- \rightleftharpoons S(s)$ $E^\circ = +1.23$ V
$V^{2+}(aq) + 2e^- \rightleftharpoons V(s)$ $E^\circ = -0.76$ V
The species with the more negative (or less positive) $E^\circ$ will be oxidized (anode), and the species with the more positive $E^\circ$ will be reduced (cathode).
Since $E^\circ_V (-0.76 \text{ V})$ is more negative than $E^\circ_S (+1.23 \text{ V})$, $V$ will be oxidized at the anode, and $S^{2+}$ will be reduced at the cathode.

Anode (oxidation): $V(s) \rightarrow V^{2+}(aq) + 2e^-$
Cathode (reduction): $S^{2+}(aq) + 2e^- \rightarrow S(s)$

The diagram should show:

• A V electrode immersed in a $V^{2+}$ solution (anode).
• An S electrode immersed in an $S^{2+}$ solution (cathode).
• A salt bridge connecting the two solutions.
• An external wire connecting the two electrodes, with an arrow indicating electron flow from V to S.
• A voltmeter in the external circuit.

        V(s) | V²⁺(aq) || S²⁺(aq) | S(s)
        (Anode)         (Cathode)

(Due to text-based format, a visual diagram cannot be drawn, but the description above outlines the key components.)

a) (iv)

The e.m.f. of the cell ($E^\circ_{cell}$) is calculated as:

From part (iii), $S$ is the cathode and $V$ is the anode.
$E^\circ_{cathode} = E^\circ_S = +1.23 \text{ V}$
$E^\circ_{anode} = E^\circ_V = -0.76 \text{ V}$

The e.m.f. of the electrochemical cell is $\boxed{\text{1.99 V}}$.