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

Options

A. standart cell potential is 0.95 v
B. during withdrawal of eelctricity from cell lead is reduced
C. during withdrawal of electrical from cell manganese is oxidized
D. during withdrawal electrical from cell two electron are transferred from lead to managese

Answer

C

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 reduced, lead should be oxidized. Reduction half reaction: `MnO_(2) +4H^(+2)e^(-) rar Mn^( +2)H_(2)O, E_(MnO_(2))^(@) = 1.23V` Oxidation half reaction: `Pb+2CI^I(-)rarr PbCI_(2),+2e^(-), E_(p//b)^(@)=-0.27V`
Cell reaction: `MnO_(2)+ Pb+2CI^(-) +4H^(+) rarr Mn^(2 +)+ PbCl_(2) +2H_(2)O, E_(cell)^(@) = 1.23+0.27-1.5 v`