Given `2H_(2)O rarr O_2 + 4H^+ + 4e^-` , `E_0 = -1.23 V`. Calculate electrode potential at pH = 5.

Explanation

To calculate the electrode potential at pH = 5 for the reaction:

\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

with a standard electrode potential \( E^0 = -1.23 \, V \), we will use the Nernst equation:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:

\[ E = E^0 - \frac{0.0591}{n} \log Q \]

where:
- \( E \) is the electrode potential,
- \( E^0 \) is the standard electrode potential,
- \( n \) is the number of electrons transferred in the reaction,
- \( Q \) is the reaction quotient.

### Step 2: Identify \( n \)
From the balanced equation, we see that 4 electrons are transferred. Therefore, \( n = 4 \).

### Step 3: Calculate the Reaction Quotient \( Q \)
The reaction quotient \( Q \) for the given reaction can be expressed in terms of the concentration of \( H^+ \):

\[ Q = [H^+]^4 \]

### Step 4: Substitute \( [H^+] \) at pH = 5
At pH = 5, the concentration of \( H^+ \) ions can be calculated using the formula:

\[ [H^+] = 10^{-\text{pH}} = 10^{-5} \, M \]

Thus,

\[ Q = (10^{-5})^4 = 10^{-20} \]

### Step 5: Substitute Values into the Nernst Equation
Now we can substitute the values into the Nernst equation:

\[ E = -1.23 - \frac{0.0591}{4} \log(10^{-20}) \]

### Step 6: Calculate \( \log(10^{-20}) \)
Using the properties of logarithms:

\[ \log(10^{-20}) = -20 \]

### Step 7: Substitute \( \log(10^{-20}) \) into the Nernst Equation
Now substituting this back into the equation:

\[ E = -1.23 - \frac{0.0591}{4} \times (-20) \]

### Step 8: Simplify the Equation
Calculating \( \frac{0.0591}{4} \):

\[ \frac{0.0591}{4} = 0.014775 \]

Now substituting this value:

\[ E = -1.23 + 0.014775 \times 20 \]

Calculating \( 0.014775 \times 20 \):

\[ 0.014775 \times 20 = 0.2955 \]

### Step 9: Final Calculation
Now substituting this back into the equation:

\[ E = -1.23 + 0.2955 = -0.9345 \, V \]

Rounding to two decimal places, we get:

\[ E \approx -0.93 \, V \]

### Final Answer
The electrode potential at pH = 5 is approximately:

\[ E \approx -0.93 \, V \]

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