Let's analyze each oxidation reaction:
A. $I^- \rightarrow I_2$:
Both $K_2Cr_2O_7$ and $KMnO_4$ can oxidize iodide to iodine in acidic medium.
This reaction is possible with both oxidizing agents.
B. $S^{2-} \rightarrow S$:
Both $K_2Cr_2O_7$ and $KMnO_4$ can oxidize sulfide to sulfur in acidic medium.
This reaction is possible with both oxidizing agents.
C. $Fe^{2+} \rightarrow Fe^{3+}$:
Both $K_2Cr_2O_7$ and $KMnO_4$ can oxidize $Fe^{2+}$ to $Fe^{3+}$ in acidic medium.
This reaction is possible with both oxidizing agents.
D. $I^- \rightarrow IO_3^-$:
Both $K_2Cr_2O_7$ and $KMnO_4$ can oxidize iodide to iodate in acidic medium, but it requires a stronger oxidizing power.
This reaction is possible under strong conditions, but not always consistent in standard acidic conditions.
E. $S_2O_3^{2-} \rightarrow SO_4^{2-}$:
Thiosulfate oxidation to sulfate is more complex and specific to certain oxidizing power and conditions. $KMnO_4$ typically does not oxidize thiosulfate to sulfate quantitatively in acidic medium, while $K_2Cr_2O_7$ may do so under vigorous conditions.
This reaction is not consistently carried out by both oxidizing agents.
Therefore, reactions A, B, and C are the only ones that are consistently carried out by both $K_2Cr_2O_7$ and $KMnO_4$ under standard acidic conditions.
Final Answer:
The final answer is $ (3)\ A,\ B\ \text{and}\ C\ \text{only} $.
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:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.