Question6. 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 halfcell?

Studdy Solution
Use the Nernst equation to calculate $\mathrm{E}_{\text{cell}}$.
First, calculate the reaction quotient $Q$:
$$Q = \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}][\mathrm{H}^{+}]^{14}}$$
Given: - $[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}] = 0.25 \, \text{M}$ - $[\mathrm{Cr}^{3+}] = 0.35 \, \text{M}$ - $\mathrm{pH} = 2.0 \Rightarrow [\mathrm{H}^{+}] = 10^{-2} \, \text{M}$
$$Q = \frac{(0.35)^2}{(0.25)(10^{-2})^{14}}$$
Now, substitute into the Nernst equation:
$$E = 1.33 - \frac{8.314 \times 298}{6 \times 96485} \ln \left( \frac{(0.35)^2}{(0.25)(10^{-2})^{14}} \right)$$
Calculate the value:
$$E = 1.33 - \frac{0.0257}{6} \ln \left( \frac{0.1225}{0.25 \times 10^{-28}} \right)$$
$$E = 1.33 - 0.00428 \ln \left( 4.9 \times 10^{26} \right)$$
$$E = 1.33 - 0.00428 \times 61.7$$
$$E = 1.33 - 0.264$$
$$E = 1.066 \, \text{V}$$
The $\mathrm{E}_{\text{cell}}$ of this half-cell is:
$$\boxed{1.066 \, \text{V}}$$