Question

calculate the $\mathrm{e^\circ_{cell}}$ for the cell given below.
$\mathrm{i_2 + 2br^- \
ightarrow 2i^- + br_2}$

half-reaction	$\mathrm{e^\circ}$ (v)

| $\mathrm{i_2 + 2e^- \
ightarrow 2i^-}$ | +0.54 |
| $\mathrm{br_2 + 2e^- \
ightarrow 2br^-}$ | +1.09 |
$\mathrm{e^\circ_{cell} = ?}$ v
enter either a + or - sign and the magnitude in the answer.

Explanation:

Step1: Identify oxidation and reduction

In the reaction \( \text{I}_2 + 2\text{Br}^-
ightarrow 2\text{I}^- + \text{Br}_2 \), \( \text{I}_2 \) gains electrons (reduction) and \( \text{Br}^- \) loses electrons (oxidation). The reduction half - reaction is \( \text{I}_2 + 2\text{e}^-
ightarrow 2\text{I}^- \) with \( E^{\circ}_{red}= + 0.54\space V \). The oxidation half - reaction is the reverse of \( \text{Br}_2 + 2\text{e}^-
ightarrow 2\text{Br}^- \), so the oxidation half - reaction is \( 2\text{Br}^-
ightarrow \text{Br}_2+ 2\text{e}^- \) and \( E^{\circ}_{ox}=-E^{\circ}_{red}=- 1.09\space V \) (since for the reverse reaction, the standard electrode potential is the negative of the forward reaction's potential).

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

The formula for the standard cell potential is \( E^{\circ}_{cell}=E^{\circ}_{red}(\text{cathode})+E^{\circ}_{ox}(\text{anode}) \). Here, the cathode is where reduction occurs (\( \text{I}_2 \) reduction) and the anode is where oxidation occurs (\( \text{Br}^- \) oxidation). So \( E^{\circ}_{cell}=E^{\circ}_{red}(\text{I}_2/2\text{I}^-)+E^{\circ}_{ox}(2\text{Br}^-/\text{Br}_2) \)
Substitute the values: \( E^{\circ}_{cell}=0.54+( - 1.09) \)
\( E^{\circ}_{cell}=0.54 - 1.09=- 0.55\space V \)

Answer:

\(-0.55\)