We are asked to **half-cell reduction potential for the anode at 25 °C.**

Recall:

$\overline{){\mathbf{}}{\mathbf{\Delta G}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{nF\epsilon}}{\mathbf{\xb0}}}$

where:

**n = number of moles electrons**

**F = Faraday’s constant, 96485 C/mol e ^{-} **

**ε° = standard reduction potential**

We can use the following equation to solve for ** ΔG˚_{rxn}**:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{prod}}{\mathbf{-}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{react}}}$

We go through the following steps to solve the problem:

**Step 1:** Write the overall chemical reaction

**Step 2:** Calculate ΔG°_{rxn}.

**Step 3:** Calculate the overall potential for the fuel cell

**Step 4:**** **Calculate half-cell reduction potential for the anode at 25 °C

**Step 1:** Write the overall chemical reaction

Recall the mnemonics **LEO GER**.

**L**ose **G**ain

**E**lectron **E**lectrons

**O**xidation **R**eduction

An illustration of an ethanol fuel cell is given below:

The half reactions (written as reductions) for the fuel cell at 25 °C are given below.

O_{2} (g) + 4 H_{3}O^{+} (aq) + 4 e^{-} → 6 H_{2}O (l) ε°_{red} = 1.23 V

CH_{3}COOH (aq) + 4 H_{3}O^{+} (aq) + 4 e^{-} → CH_{3}CH_{2}OH (aq) + 5 H_{2}O (l) ε°_{red} = ???

* acetic acid ethanol*

Calculate the half-cell reduction potential for the anode at 25 °C. Please circle your answer.

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