Q.Indentify the strongest and the weakest oxidizing agents from the electrochemical series.

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.

Given: A = -3.05, B = -1.66, C = -0.40, D = +0.80 V.

Most negative = A (-3.05 V), so A has the highest reducing power (like an alkali metal).

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:

• (a) The EMF is not negative — it is precisely because it is positive that the corrosion happens; a negative EMF would mean the reaction is non-spontaneous and the zinc vessel would in fact be safe.

• (c) False statement — the reduction potential of $Zn$ ($-0.76\ V$) is actually less than (more negative than) that of $Cu$ ($+0.34\ V$), not more.

• (d) False statement — the reduction potential of $Cu$ is $+0.34\ V$, which is positive, not negative.

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.

Given values: A = -3.05 V, B = -1.66 V, C = -0.40 V, D = +0.80 V. The most negative is A (-3.05 V), so metal A has the greatest tendency to be oxidised and the highest reducing power. (D, being most positive, is the weakest reducing agent / strongest oxidising agent.)

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)
• B: E° = -2.50 V (most negative → strongest reducing agent)
• C: E° = -0.50 V (intermediate)

Arranging in decreasing order of reducing power (most negative first):

B (-2.50 V) > C (-0.50 V) > A (+0.68 V)

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 $E^\circ$.

Given $E^\circ$: $A=+1.5\,V,\ B=-2.0\,V,\ C=+0.34\,V,\ D=-0.76\,V$.

Most reactive = most negative: $B(-2.0)>D(-0.76)>C(+0.34)>A(+1.5)$.

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

A species with a more negative standard reduction potential is more easily oxidised (a stronger reducing agent), while a species with a higher (more positive) reduction potential is more easily reduced (a stronger oxidising agent). Here $E^\circ(A^+/A) = -1.66\ V$ (very negative, so $A$ is readily oxidised to $A^+$) and $E^\circ(B^+/B) = +0.80\ V$ (so $B^+$ is readily reduced to $B$). The spontaneous combination is therefore $A$ being oxidised while $B^+$ is reduced:

$A + B^+ \rightarrow A^+ + B$

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.

Comparing the four values: A = -3.06 V, B = -1.63 V, C = -0.40 V, D = -0.80 V. The most negative is A (-3.06 V), so A is the most easily oxidised and the strongest reducing agent.