In acidic medium, only iodide ions (I\(^-\)) can be oxidized by oxygen to form iodine (\( \text{I}_2 \)). The reaction is as follows: \[ 4 \text{I}^- + 4 \text{H}^+ + \text{O}_2 \rightarrow 2 \text{I}_2 (s) + 2 \text{H}_2\text{O} (l). \] Chloride (Cl\(^-\)) and bromide (Br\(^-\)) ions are not oxidized by oxygen in acidic medium because their reduction potentials are not favorable for such oxidation under standard conditions.
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).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: