Consider the standard reduction potentials (E^\circ) at 25^\circ\text{C} for the following… — JAMB UTME

Consider the standard reduction potentials ( $E^\circ$ ) at $25^\circ\text{C}$ for the following half-reactions:

\text{Ce}^{4+}(aq) + e^- \rightarrow \text{Ce}^{3+}(aq) \quad E^\circ = +1.61\text{ V}

\text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq) \quad E^\circ = +1.07\text{ V}

\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}

\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \quad E^\circ = +0.54\text{ V}

\text{Sn}^{4+}(aq) + 2e^- \rightarrow \text{Sn}^{2+}(aq) \quad E^\circ = +0.15\text{ V}

Which of the following chemical species can spontaneously oxidize $\text{I}^-(aq)$ to $\text{I}_2(s)$ under standard conditions, but is UNABLE to oxidize $\text{Br}^-(aq)$ to $\text{Br}_2(l)$ ?

$\text{Fe}^{3+}(aq)$ Answer
B
$\text{Ce}^{4+}(aq)$
C
$\text{Sn}^{4+}(aq)$
D
$\text{Sn}^{2+}(aq)$

Answer

The species

\text{Fe}^{3+}(aq)

is the correct choice because its standard reduction potential (+0.77 V) lies strictly between the reduction potentials of

\text{I}_2/\text{I}^-

(+0.54 V) and

\text{Br}_2/\text{Br}^-

(+1.07 V).

To oxidize

\text{I}^-

(

E^\circ_{\text{red}} = +0.54\text{ V}

), an oxidizing agent must have a standard reduction potential greater than

+0.54\text{ V}

. To fail to oxidize

\text{Br}^-

(

E^\circ_{\text{red}} = +1.07\text{ V}

), its reduction potential must be less than

+1.07\text{ V}

. The species

\text{Fe}^{3+}(aq)

has

E^\circ = +0.77\text{ V}

, which satisfies

+0.54\text{ V} < +0.77\text{ V} < +1.07\text{ V}

. Thus,

E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{I}^-) = +0.77 - 0.54 = +0.23\text{ V} > 0

(spontaneous) and

E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{Br}^-) = +0.77 - 1.07 = -0.30\text{ V} < 0

(non-spontaneous).

Step-by-Step Solution

Determine the required condition for spontaneous oxidation of

\text{I}^-(aq)

\text{I}_2(s)

The oxidation half-reaction is

2\text{I}^-(aq) \rightarrow \text{I}_2(s) + 2e^-

with

E^\circ_{\text{ox}} = -0.54\text{ V}

. For

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} > 0

, the oxidant must have

E^\circ_{\text{red}} > +0.54\text{ V}

A redox reaction is spontaneous if the cell potential

E^\circ_{\text{cell}}

is positive.

Determine the required condition for non-spontaneous oxidation of

\text{Br}^-(aq)

\text{Br}_2(l)

The oxidation half-reaction is

2\text{Br}^-(aq) \rightarrow \text{Br}_2(l) + 2e^-

with

E^\circ_{\text{ox}} = -1.07\text{ V}

. For

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} < 0

, the oxidant must have

E^\circ_{\text{red}} < +1.07\text{ V}

An unfeasible (non-spontaneous) reaction corresponds to a negative overall standard cell potential.

Combine the potential boundary constraints and evaluate the options.

The standard reduction potential of the ideal oxidant must satisfy

+0.54\text{ V} < E^\circ_{\text{red}} < +1.07\text{ V}

. Among the choices,

\text{Fe}^{3+}(aq)

has

E^\circ_{\text{red}} = +0.77\text{ V}

, which falls squarely within this range.

Comparing standard electrode potentials directly establishes which species can act as selective oxidizing agents.

Key Concept

Predicting reaction spontaneity and selective oxidation using standard electrode potentials (

E^\circ

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic