Question

calculate the e°cell value for the cell given below.
mn | mn²⁺ || fe³⁺, fe²⁺ | pt
half-reaction | e° (v)
mn²⁺ + 2e⁻ → mn | -1.18
fe³⁺ + e⁻ → fe²⁺ | +0.77
e°cell = ? v
enter either a + or - sign and the magnitude in the answer.

Explanation:

Step1: Identify anode and cathode

Anode (oxidation): Mn → Mn²⁺ + 2e⁻ (reverse of Mn²⁺ + 2e⁻ → Mn), so \( E^{\circ}_{\text{anode}} = +1.18 \) V (reverse the given value for oxidation).
Cathode (reduction): Fe³⁺ + e⁻ → Fe²⁺, \( E^{\circ}_{\text{cathode}} = +0.77 \) V.

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

Use the formula \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode (reduction)}} \) or directly \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode (reduction)}} + E^{\circ}_{\text{anode (oxidation)}} \).
Substitute values: \( E^{\circ}_{\text{cell}} = 0.77 + 1.18 = 1.95 \) V.

Answer:

+1.95