Reduction Half Reaction:
\[ 2\text{MnO}_4^- \rightarrow 2\text{MnO}_2 \] \[ 2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- \]
Oxidation Half Reaction:
\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \] \[ 6\text{I}^- \rightarrow 3\text{I}_2 + 6e^- \]
Adding the oxidation half and reduction half, we get the net reaction as:
\[ 2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 3\text{I}_2 + 2\text{MnO}_2 + 8\text{OH}^- \]
Thus, \( z = 8 \).
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32