JEE MAIN - Chemistry (2022 - 25th June Evening Shift - No. 7)
The correct order of reduction potentials of the following pairs is
A. Cl2/Cl$$-$$
B. I2/I$$-$$
C. Ag+/Ag
D. Na+/Na
E. Li+/Li
Choose the correct answer from the options given below.
A > C > B > D > E
A > B > C > D > E
A > C > B > E > D
A > B > C > E > D
Explanation
$$\mathrm{E}_{\mathrm{C}_{2} / \mathrm{CI}}^{\circ}=+1.36 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{0}=+0.54 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\circ}=-2.71 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{L}^{+} / \mathrm{Li}}=-3.05 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{0}=+0.54 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\circ}=-2.71 \mathrm{~V}$$
$$\mathrm{E}_{\mathrm{L}^{+} / \mathrm{Li}}=-3.05 \mathrm{~V}$$
Comments (0)
