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

In the cell notation \( \text{Mn} \mid \text{Mn}^{2+} \parallel \text{Fe}^{3+}, \text{Fe}^{2+} \mid \text{Pt} \), the left side (Mn) is the anode (oxidation) and the right side (Fe) is the cathode (reduction).

For the anode (oxidation), reverse the half - reaction of \( \text{Mn}^{2+}+2e^-
ightarrow\text{Mn} \). The oxidation half - reaction is \( \text{Mn}
ightarrow\text{Mn}^{2+}+2e^- \), and its \( E^{\circ}_{\text{ox}}=-E^{\circ}_{\text{red}} = -(- 1.18)=1.18\space V \) (since the given \( E^{\circ} \) for \( \text{Mn}^{2+}+2e^-
ightarrow\text{Mn} \) is \( - 1.18\space V \)).

For the cathode (reduction), the half - reaction is \( \text{Fe}^{3+}+e^-
ightarrow\text{Fe}^{2+} \) with \( E^{\circ}_{\text{red}} = 0.77\space V \).

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

The formula for \( E^{\circ}_{\text{cell}} \) is \( E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode (reduction)}}+E^{\circ}_{\text{anode (oxidation)}} \)

Substitute the values: \( E^{\circ}_{\text{cell}}=0.77 + 1.18 \)

Step3: Compute the Result

\( 0.77+1.18 = 1.95\space V \)

Answer:

\( +1.95 \)