Question

half-reaction | e° (v)
a⁺ + e⁻ → a | -0.5
b⁺ + e⁻ → b | 0.0
c⁺ + e⁻ → c | +1.0
what is the e°_cell value for the spontaneous reaction between substance a and substance b?
-0.5 v
1.5 v
+0.5 v
-1.5 v

Explanation:

Step1: Determine Oxidation and Reduction

For a spontaneous reaction, \( E^{\circ}_{\text{cell}}>0 \). We need to identify which species is oxidized (loses electrons) and which is reduced (gains electrons). The half - reactions are:

• \( \text{A}^++\text{e}^-

ightarrow\text{A} \), \( E^{\circ}=- 0.5\space V \)

• \( \text{B}^++\text{e}^-

ightarrow\text{B} \), \( E^{\circ}=0.0\space V \)

A species with a more negative (or less positive) reduction potential will be oxidized (reverse the half - reaction). So, A will be oxidized (since its reduction potential is more negative than B's). The oxidation half - reaction is \( \text{A}
ightarrow\text{A}^++\text{e}^- \), and the reduction potential for the reverse (oxidation) is \( E^{\circ}_{\text{ox}} = 0.5\space V \) (reverse of \( \text{A}^++\text{e}^-
ightarrow\text{A} \), so we change the sign of \( E^{\circ} \)).

B will be reduced: \( \text{B}^++\text{e}^-
ightarrow\text{B} \), \( E^{\circ}_{\text{red}}=0.0\space V \)

Step2: Calculate \( E^{\circ}_{\text{cell}} \)

The formula for \( E^{\circ}_{\text{cell}} \) is \( E^{\circ}_{\text{cell}}=E^{\circ}_{\text{red}}+E^{\circ}_{\text{ox}} \) (or \( E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \), where cathode is the reduction half - cell and anode is the oxidation half - cell).

Using \( E^{\circ}_{\text{cell}}=E^{\circ}_{\text{red (cathode)}}+E^{\circ}_{\text{ox (anode)}} \), here \( E^{\circ}_{\text{red (B)}} = 0.0\space V \) and \( E^{\circ}_{\text{ox (A)}}=0.5\space V \) (since oxidation of A is reverse of its reduction, so \( E^{\circ}_{\text{ox}}=-E^{\circ}_{\text{red (A)}}=0.5\space V \)).

So \( E^{\circ}_{\text{cell}}=0.0 + 0.5= + 0.5\space V \)

Answer:

+0.5 V