The half reaction with their oxidation potential are Pb(s)rarrPb^(2+)(aq)+2 e^(-)E_("xi")^(@)=+0.13 V, Ag (s) rarr (aq)^(+)(aq)+e^(-),E_("xi")^(@)=-0.80 V write the cell reaction and calcualte its EMF

The half reaction with their oxidation potential are Pb(s)rarrPb^(2+)

1.	The half reaction with their oxidation potential are Pb(s)rarrPb^(2+)(aq)+2 e^(-)E_("xi")^(@)=+0.13 V, Ag (s) rarr (aq)^(+)(aq)+e^(-),E_("xi")^(@)=-0.80 V write the cell reaction and calcualte its EMF
Answer» Solution :Rewrite the two equatin in the reduction form by reversing the sings of oxidatin potention thus `Pb^(2+)(AQ)+2 e^(-) rarr Pb(s) , E^(@)=-0.13V` `Ag^(+)(aq)+e^(-)rarr Ag(s),E^(@)=+0.80V` To obtain the equation for the cell reaction MULTIPLY Eq (ii) with 2 and SUBTRACT Eq (i) from it we have `Pb(s)+2 Ag^(+)(aq) rarr Pb^(2+)(aq)+2AG(s),E_(cell)^(@)=-0.80 -(0.13)=+0.93 V`