Q.vi. Consider the half reactions with standard potentials i. Ag$^+$ (aq) + e$^-$ $\longrightarrow$ Ag (s)$\quad E^0$ = 0.8V ii. I$_2$ (s) + 2e$^-$ $\longrightarrow$ 2I$^-$ (aq)$\quad E^0$ = 0.53V iii. Pb$^{2+}$ (aq) + 2e$^-$ $\longrightarrow$ Pb (s)$\quad E^0$ = -0.13V iv. Fe$^{2+}$ (aq) + 2e$^-$ $\longrightarrow$ Fe (s)$\quad E^0$ = - 0.44V The strongest oxidising and reducing agents respectively are a. Ag and Fe$^{2+}$
b. Ag$^+$ and Fe
c. Pb$^{2+}$ and I$^-$
d. I$_2$ and Fe$^{2+}$

The more negative the standard reduction potential, the stronger the tendency to be oxidised, i.e. the greater the reducing power. A has -3.05 V, the most negative.

A species with a very negative reduction potential is easily oxidised (gives up electrons readily), so it is a strong reducing agent. …

Zinc's much lower (more negative) standard reduction potential than copper's makes the displacement reaction's EMF positive, i.e. spontaneous — which is exactly why $Zn$ metal corrodes when it touches $CuSO_4$ solution.

Standard reduction potentials: $E^{\circ}(Cu^{2+}/Cu) = +0.34\ V$; $E^{\circ}(Zn^{2+}/Zn) = -0.76\ V$. Since $Zn^{2+}/Zn$ has the far more negative (lower) reduction potential, $Zn$ metal is the stronger reducing agent and is oxidised preferentially:

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34-(-0.76) = +1.10\ V$

Since $\Delta G^{\circ} = -nFE^{\circ}_{cell}$, a positive $E^{\circ}_{cell}$ gives a negative $\Delta G^{\circ}$ — the reaction is thermodynamically spontaneous. So $CuSO_4$ solution spontaneously reacts with a zinc container, dissolving the zinc (corroding it) and depositing copper metal on it.

Why the other options are wrong: …

The more negative the standard reduction potential, the stronger the reducing agent — so A (-3.05 V) has the highest reducing power.

Reducing power measures the tendency of a metal to lose electrons (get oxidised). A more negative standard electrode (reduction) potential means the metal is more easily oxidised and is therefore a stronger reducing agent.

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Lower (more negative) standard reduction potential means a stronger tendency to be oxidised, i.e. a stronger reducing agent.

Reducing power measures how readily a species gives up electrons (undergoes oxidation). The more negative the standard reduction potential (E°), the greater the tendency to lose electrons, so the stronger the reducing agent.

Given:

• A: E° = +0.68 V (least negative → weakest reducing agent) …

Lower (more negative) $E^\circ$ ⇒ stronger reducing agent ⇒ more reactive metal; so $B>D>C>A$.

Concept: A metal's chemical activity (its tendency to lose electrons and get oxidised) increases as its standard reduction potential becomes more negative. Metals high in the activity series (like the alkali/alkaline-earth metals) have strongly negative $E^\circ$.

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$A$ has the more negative (lower) reduction potential, so it is the better reducing agent and is oxidised, while $B^+$ (higher reduction potential) is reduced.

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Most negative reduction potential = strongest tendency to be oxidised = strongest reducing agent, i.e. A (-3.06 V).

A reducing agent works by giving up electrons (being oxidised). The more negative the standard reduction potential, the greater the tendency of that species to undergo oxidation rather than reduction, and hence the stronger it is as a reducing agent.

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