Act like a helpful tutor and exlain step-by-step solution with explanation and final answer:321 Electromotive force
Consider 3 polymer electrolyte fuel cell using. #*-conducting electrolyte, compused

Question

Act like a helpful tutor and exlain step-by-step solution with explanation and final answer:321 Electromotive force
Consider 3 polymer electrolyte fuel cell using. #*-conducting electrolyte, compused of the
following cell construction. Note that po, fatm] denotes the oxygen gas partial pressure and
Gaon Tepresents the seater activity with respect to pre water ds the standird stare.
Hag), w OREO.
Puy =100 fam] | Polymer electrolye | 70: 0205 atm)
analy =1
(1) Determine the half=cell reactions of the efe and right electrodes of this battery, and obtain
the overall full-cel reaction equition.
(2) Based an the thetmochemical dita in Table 1, caleulite the standard clectromative force
Ey [V) and apen circuit voltage (equilibrium electromative force) EV] of this cell at
298 K up to 3 significant figures.

VIPSolutions · Accepted Answer

Let's break down the problem step by step.

---

## **Step 1: Write the Half-Cell Reactions**

### **Anode (Left Electrode):**
- **Hydrogen oxidation:**  
  $\text{H}_2(\text{g}) \rightarrow 2\text{H}^+ + 2e^-$  
  *At Pt, $p_{H_2} = 1.00 \text{ atm}$*

### **Cathode (Right Electrode):**
- **Oxygen reduction:**  
  $\frac{1}{2}\text{O}_2(\text{g}) + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}(\ell)$  
  *At Pt, $p_{O_2} = .205 \text{ atm},\ a_{H_2O(\ell)} = 1$*

---

## **Step 2: Write the Overall Cell Reaction**

Add the two half-cell reactions (electrons will cancel):

$$
\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\ell)
$$

---

## **Step 3: Standard Electromotive Force ($E^$) Calculation**

From standard electrode potentials (at 298 K, all gases at 1 atm, solutes at 1 M):

- $E^_{\text{cathode}} (\text{O}_2/\text{H}_2\text{O}) = +1.229 \text{ V}$
- $E^_{\text{anode}} (\text{H}_2/\text{H}^+) =  \text{ V}$

$$
E^_{\text{cell}} = E^_{\text{cathode}} - E^_{\text{anode}} = 1.229\ \text{V} - \ \text{V} = \boxed{1.23\ \text{V}}
$$

---

## **Step 4: Nernst Equation for Non-Standard Conditions**

The cell operates at:
- $p_{H_2} = 1.00 \text{ atm}$
- $p_{O_2} = .205 \text{ atm}$
- $a_{H_2O(\ell)} = 1$

The Nernst equation for the cell is:

$$
E = E^_{\text{cell}} - \frac{.05916}{n} \log Q
$$

Where:
- $n = 2$ (electrons transferred)
- $Q = \frac{a_{H_2O}}{p_{H_2} \cdot (p_{O_2})^{1/2}}$

Plug in values:

$$
Q = \frac{1}{1.00 \times (.205)^{1/2}}
$$

$$
(.205)^{1/2} \approx .453
$$

$$
Q = \frac{1}{1 \times .453} = 2.21
$$

---

$$
E = 1.229 - \frac{.05916}{2} \log(2.21)
$$

$$
\log(2.21) \approx .344
$$

$$
\frac{.05916}{2} = .02958
$$

$$
E = 1.229 - (.02958 \times .344)
$$
$$
E = 1.229 - .0102
$$
$$
E = \boxed{1.22\ \text{V}}
$$

---

## **Final Answers**

### **(1) Half-cell reactions and overall cell reaction:**
- **Anode:** $\text{H}_2(\text{g}) \rightarrow 2\text{H}^+ + 2e^-$
- **Cathode:** $\frac{1}{2}\text{O}_2(\text{g}) + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{O}(\ell)$
- **Overall:** $\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\ell)$

### **(2) Standard EMF and cell voltage at given conditions:**
- **Standard EMF ($E^$):** $\boxed{1.23\ \text{V}}$
- **Open circuit voltage at given conditions:** $\boxed{1.22\ \text{V}}$ (to 3 significant figures)

Let me know if you need more detail on any step!

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