Math / Calculus

Question5. Calculate the activity coefficients of sodium and chloride ions for a 0.04 M NaCl solution
6. What is an electrode of the first kind? Write a general Nernst equation for this indicator electrode.
7. A solution of $0.25 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ and $0.35 \mathrm{M} \mathrm{Cr}^{3+}$ has a pH of 2.0 . What is the $\mathrm{E}_{\text {cell }}$ of this half-cell?
8. In an acidic solution, $\mathrm{O}_{2}(\mathrm{~g})$ oxidizes $\mathrm{Cr}^{2+}(\mathrm{aq})$ to $\mathrm{Cr}^{3+}(\mathrm{aq})$ . The $\mathrm{O}_{2}(\mathrm{~g})$ is reduced to $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ . $E_{\text {cell }}^{o}$ for this reaction is +1.653 V . What is the electrode potential for the $\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}$ half-cell?

Studdy Solution

STEP 1

1. The reaction involves $\text{O}_2$ oxidizing $\text{Cr}^{2+}$ to $\text{Cr}^{3+}$ .
2. The standard cell potential $E_{\text{cell}}^{o}$ for the overall reaction is $+1.653 \, \text{V}$ .

STEP 2

1. Write the half-reactions.
2. Use the standard cell potential to find the unknown half-cell potential.
3. Apply the Nernst equation if needed.

STEP 3

Write the half-reactions for the overall reaction:
- Oxidation: $\text{Cr}^{2+} \rightarrow \text{Cr}^{3+} + e^-$ - Reduction: $\text{O}_2 + 4H^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$

STEP 4

Use the given standard cell potential to find the potential of the $\text{Cr}^{3+}/\text{Cr}^{2+}$ half-cell.
The overall cell potential is given by:
$E_{\text{cell}}^{o} = E_{\text{red}}^{o}(\text{O}_2/\text{H}_2\text{O}) - E_{\text{ox}}^{o}(\text{Cr}^{3+}/\text{Cr}^{2+})$
Given $E_{\text{cell}}^{o} = +1.653 \, \text{V}$ .
Assume $E_{\text{red}}^{o}(\text{O}_2/\text{H}_2\text{O})$ is known from standard tables (approximately $+1.23 \, \text{V}$ ).
$1.653 = 1.23 - E_{\text{ox}}^{o}(\text{Cr}^{3+}/\text{Cr}^{2+})$
Solve for $E_{\text{ox}}^{o}(\text{Cr}^{3+}/\text{Cr}^{2+})$ :
$E_{\text{ox}}^{o}(\text{Cr}^{3+}/\text{Cr}^{2+}) = 1.23 - 1.653$
$E_{\text{ox}}^{o}(\text{Cr}^{3+}/\text{Cr}^{2+}) = -0.423 \, \text{V}$
The electrode potential for the $\text{Cr}^{3+}/\text{Cr}^{2+}$ half-cell is:
$\boxed{-0.423 \, \text{V}}$

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