Thehalf cel reactions with their oxidation potentials are
(a) Pb(s)`-2e^(-)toPb^(2+)(aq),E_("oxi")^(@)=+0.13V` (b) `Ag(s)-e^(-)toAg^(+)(aq),E_("oxi")^(@)=0.80V`
Write the cell reaction and calculate its emf.

Explanation

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`