Consider the half reactions of a galvanic cell given below: `MnO_(2) + H^(+) hArr Mn^(2+) + 2H_(2)O, E_(MnO_(2))^(@)= 1.23V, PbCl_(2) hArr Pb+ 2Cl^(-), E_(PbCl_(2)^(@)= -0.27V`. The correct statement about the cell is

Options

A. Standard cell potential is 0.95V
B. During withdrawal of electricity from cell, lead is reduced
C. During withdrawal of electricity from cell, manganese is oxidized.
D. During withdrawal of electricity from cell, two electrons are transferred from lead to manganese.

Answer

D

Explanation

The reduction potential of `MnO_(2)` is more than that of `PbCl_(2)`. So, `Pb^(2+)` cannot reduce manganese oxide. For cell reaction to occur, redox reaction is necessary. Hence, if `MnO_(2)` is reducecd, lead should be oxidized. Reduction half reaction: `MnO_(2) + 4H^(+) + 2e^(-) hArr Mn^(2+) + 2H_(2)O, E_(MnO_(2))^(@)= 1.23V`
Oxidation half reaction : `Pb+ 2Cl^(-) hArr PbCl_(2) + 2e^(-), E_(Pb)^(@)= -0.27V`
Cell reaction: `MnO_(2) + Pb+ 2Cl^(-) + 4H^(+) hArr Mn^(2+) + PbCl_(2) + 2H_(2)O`
`E_("cell")^(@)= 1.23 + 0.27= 1.5V`