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 c?
+0.5 v
1.5 v
-1.5 v
-0.5 v

Explanation:

Step1: Identify Oxidation and Reduction

For a spontaneous reaction, \( E^{\circ}_{\text{cell}}>0 \). Oxidation is loss of electrons, reduction is gain.

• A: \( \text{A}^+ + e^-

ightarrow \text{A} \) (reduction, \( E^{\circ}_{\text{red}} = -0.5 \, \text{V} \)) – but for oxidation, reverse it: \( \text{A}
ightarrow \text{A}^+ + e^- \), so \( E^{\circ}_{\text{ox}} = +0.5 \, \text{V} \) (since oxidation is reverse of reduction, \( E^{\circ}_{\text{ox}}=-E^{\circ}_{\text{red}} \)).

• C: \( \text{C}^+ + e^-

ightarrow \text{C} \) (reduction, \( E^{\circ}_{\text{red}} = +1.0 \, \text{V} \)).

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

\( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red (cathode)}} + E^{\circ}_{\text{ox (anode)}} \).
Cathode (reduction): \( \text{C}^+ + e^-
ightarrow \text{C} \) (\( E^{\circ}_{\text{red}} = +1.0 \, \text{V} \)).
Anode (oxidation): \( \text{A}
ightarrow \text{A}^+ + e^- \) (\( E^{\circ}_{\text{ox}} = +0.5 \, \text{V} \)).

Thus, \( E^{\circ}_{\text{cell}} = 1.0 + 0.5 = 1.5 \, \text{V} \).

Answer:

1.5 V