The table gives the standard redox potentials for a number of half-reactions. (i) Relative to which half-cell reaction are the above electrode potentials expressed? (ii) Calculate the e.m.f of the cel

The table gives the standard redox potentials for a number of half-reactions:

Standard electrode potentials are always measured relative to the standard hydrogen electrode (SHE).
The electrode potentials are expressed relative to the $\boxed{\text{Standard Hydrogen Electrode (SHE)}}$.

I. The strongest oxidising agent
The strongest oxidizing agent is the species that is most easily reduced (has the most positive standard reduction potential).
Comparing the E° values:

• Zn²⁺: $-0.76 \text{ V}$
• Fe²⁺: $-0.44 \text{ V}$
• I₂: $+0.54 \text{ V}$
• Fe³⁺: $+0.77 \text{ V}$
• Ag⁺: $+0.88 \text{ V}$ The most positive E° is $+0.88 \text{ V}$, corresponding to $Ag^+(aq)$. The strongest oxidizing agent is $\boxed{Ag^+(aq)}$.

II. The strongest reducing agent
The strongest reducing agent is the species that is most easily oxidized (has the most negative standard reduction potential).
Comparing the E° values:

• Zn(s): $-0.76 \text{ V}$
• Fe(s): $-0.44 \text{ V}$
• 2I⁻(aq): (from $I_2(l) + 2e^- \to 2I^-(aq)$)
• Fe²⁺(aq): (from $Fe^{3+}(aq) + e^- \to Fe^{2+}(aq)$)
• Ag(s): (from $Ag^+(aq) + e^- \to Ag(s)$) The most negative E° is $-0.76 \text{ V}$, corresponding to $Zn(s)$. The strongest reducing agent is $\boxed{Zn(s)}$.

To convert iodide ions ($I^-$) to iodine ($I_2$), the iodide ions must be oxidized. The half-reaction for this oxidation is:

The standard oxidation potential for this reaction is the negative of the standard reduction potential for $I_2(l) + 2e^- \to 2I^-(aq)$, so $E^\circ_{oxidation} = -0.54 \text{ V}$.
For a spontaneous reaction, we need an oxidizing agent (a species that gets reduced) with a standard reduction potential ($E^\circ_{reduction}$) that is more positive than $+0.54 \text{ V}$ (the reduction potential of $I_2$).

From the table, the substances with $E^\circ > +0.54 \text{ V}$ are:

• $Fe^{3+}(aq)$ ($E^\circ = +0.77 \text{ V}$)
• $Ag^+(aq)$ ($E^\circ = +0.88 \text{ V}$)

Therefore, $Fe^{3+}(aq)$ and $Ag^+(aq)$ can be used to convert iodide ions to iodine.

Possible conversions:

1. Using $Fe^{3+}(aq)$: Reduction: $Fe^{3+}(aq) + e^- \to Fe^{2+}(aq)$ Oxidation: $2I^-(aq) \to I_2(l) + 2e^-$ To balance electrons, multiply the reduction half-reaction by 2:

   $$2Fe^{3+}(aq) + 2e^- \to 2Fe^{2+}(aq)$$

   Adding the half-reactions:

   $$\boxed{2Fe^{3+}(aq) + 2I^-(aq) \to 2Fe^{2+}(aq) + I_2(l)}$$

2. Using $Ag^+(aq)$: Reduction: $Ag^+(aq) + e^- \to Ag(s)$ Oxidation: $2I^-(aq) \to I_2(l) + 2e^-$ To balance electrons, multiply the reduction half-reaction by 2:

   $$2Ag^+(aq) + 2e^- \to 2Ag(s)$$

   Adding the half-reactions:

   $$\boxed{2Ag^+(aq) + 2I^-(aq) \to 2Ag(s) + I_2(l)}$$

The standard electrode potential for the elements chlorine and magnesium are:-
$Cl_2(g) + 2e^- \to 2Cl^-(aq) \quad E^\circ = +1.36 \text{ V}$
This part of the question is incomplete as it only provides the standard electrode potential for chlorine and states that it will also provide it for magnesium, but no question is asked.