On charging the lead storage battery, the oxidation state of lead changes from $\mathrm{x}_{1}$ to $\mathrm{y}_{1}$ at the anode and from $\mathrm{x}_{2}$ to $\mathrm{y}_{2}$ at the cathode. The values of $\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{x}_{2}, \mathrm{y}_{2}$ are respectively:
To solve this question, we need to understand the oxidation states involved in the charging process of a lead storage battery. Let's break down the process at both the anode and cathode and calculate the changes in oxidation states:
Therefore, the correct values for the oxidation state changes are:
Thus, the correct option for the values of \(x_1, y_1, x_2, y_2\) is:
$+2, 0, +2, +4$
1. Anode reaction: - $\mathrm{PbSO}_{4}$ is reduced to $\mathrm{Pb}$. - $\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{0}$ - $\mathrm{x}_{1} = +2$, $\mathrm{y}_{1} = 0$
2. Cathode reaction: - $\mathrm{PbSO}_{4}$ is oxidized to $\mathrm{PbO}_{2}$. - $\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+}$ - $\mathrm{x}_{2} = +2$, $\mathrm{y}_{2} = +4$
Therefore, the correct answer is (2) $+2,0,+2,+4$.


Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.