Write the given standard reduction potentials

The given half-reactions are written as oxidations, but the standard electrode potentials provided correspond to the reduction directions:

$Fe^{3+} + e^- \to Fe^{2+}, \qquad E^\circ = 0.77\,\text{V}$

$I_2 + 2e^- \to 2I^-, \qquad E^\circ = 0.54\,\text{V}$

Decide which reaction will be reduction and which will be oxidation

In a spontaneous galvanic cell:

• the half-cell with higher reduction potential undergoes reduction at cathode,
• the other undergoes oxidation at anode.

Since

$0.77\,\text{V} > 0.54\,\text{V},$

we have:

• Cathode (reduction): $Fe^{3+} + e^- \to Fe^{2+}$
• Anode (oxidation): reverse of iodine reduction, $2I^- \to I_2 + 2e^-$

Balance electrons

Multiply the iron half-reaction by 2:

$2Fe^{3+} + 2e^- \to 2Fe^{2+}$

Add with

$2I^- \to I_2 + 2e^-$

Net reaction:

$2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$

Calculate standard cell potential

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

Here the anode value is taken as the reduction potential of the iodine couple:

$E^\circ_{cell} = 0.77 - 0.54 = 0.23\,\text{V}$

Match with the form given

We need

$E^\circ_{cell} = x \times 10^{-2}\,\text{V}$

Since

$0.23\,\text{V} = 23 \times 10^{-2}\,\text{V},$

therefore,

$x = 23$

Compare with stored answer

Stored correct answer = 23, which matches our result.