For the gaseous reaction \( \text{N}_2\text{O}_5 \rightarrow 2\text{NO}_2 + \frac{1}{2}\text{O}_2 \), the rate can be expressed as follows:

The correct relation between \( K_1, K_2 \) and \( K_3 \) is
\( 4k_1 = 2k_2 = k_3 \)
The rate of disappearance of \( \text{N}_2\text{O}_5 \) is \( -\frac{d[\text{N}_2\text{O}_5]}{dt} = k_1[\text{N}_2\text{O}_5] \),
the formation of \( \text{NO}_2 \) is \( \frac{d[\text{NO}_2]}{dt} = 2k_1[\text{N}_2\text{O}_5] \),
and the formation of \( \text{O}_2 \) is \( \frac{d[\text{O}_2]}{dt} = \frac{1}{2}k_1[\text{N}_2\text{O}_5] \).
Balancing the coefficients, we find \( 2k_1 = k_2 = 4k_3 \).
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)