Question

Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq). 

Hint:

Use the Nernst equation to adjust the standard potential based on the concentration of the species involved in the reaction.

Answer 1

Step 1: The Nernst Equation 

The Nernst equation is given by: \[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \] Where:

• \( E^\circ \) is the standard electrode potential,
• \( n \) is the number of electrons involved in the reaction,
• The concentrations of products and reactants are represented in molarity.

Step 2: Substituting the Given Values

We are given:

• \( E^\circ = 1.33 \, \text{V} \),
• \( n = 6 \),
• \( [\text{Cr}^{3+}] = 0.01 \, \text{M} \),
• \( [\text{H}_2\text{O}] = 1 \, \text{M} \) (liquid, so it is 1 M),
• \( [\text{Cr}_2\text{O}_7^{2-}] = 0.01 \, \text{M} \),
• \( [\text{H}^+] = 1.0 \times 10^{-4} \, \text{M} \).

Substituting these values into the Nernst equation: \[ E = 1.33 - \frac{0.0591}{6} \log \left( \frac{(0.01)^2 \times 1}{(0.01) \times (1.0 \times 10^{-4})^{14}} \right) \]

 

Step 3: Simplifying the Logarithmic Term

Simplifying the expression inside the logarithm: \[ \frac{(0.01)^2 \times 1}{(0.01) \times (1.0 \times 10^{-4})^{14}} = \frac{0.0001}{0.01 \times (1.0 \times 10^{-4})^{14}} = \frac{0.0001}{10^{-58}} = 10^{54} \]

Step 4: Substituting the Result into the Nernst Equation

Now substitute \( 10^{54} \) back into the Nernst equation: \[ E = 1.33 - \frac{0.0591}{6} \log(10^{54}) \] Since \( \log(10^{54}) = 54 \), we get: \[ E = 1.33 - \frac{0.0591}{6} \times 54 \]

Step 5: Calculate the Final Value of \( E \)

Performing the multiplication: \[ \frac{0.0591 \times 54}{6} = 0.5319 \] So: \[ E = 1.33 - 0.5319 = 0.7981 \, \text{V} \]

Final Answer:

The final value of \( E \) is approximately: \[ E = 0.798 \, \text{V} \]