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
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:
$$\boxed{-0.423 \, \text{V}}$$