The half cell reactions with reduction potentials are
`Pb(s) to Pb^(+2) (aq), E_("red")^(@) =-0.13 V`
`Ag(s) to Ag^(+) (aq) , E_("rod")^(@) =+0.80 V`
Calculate its emf.

Explanation

The reaction potentials of the two half cell reactions are :
`Pb^(2+) (aq) + 2e^(-) to Pb (s) , (E^(Theta) = -0.13 V)`
`Ag^(+) (aq) + e^(-) to Ag (s) , (E^(Theta) = + 0.80 V)`
The reduction potential of `Ag^(+)|Ag` electrode is more than that of `Pb^(2+) |Pb` . Therefore , reduction will occur at silver electrode and oxidation will occur at lead electrode and the cell reaction will be
`2 Ag^(+) (aq) + Pb (s) to 2 Ag (s) + Pb^(2+) (aq)`
The cell may be represented as :
`Pb |Pb^(2+) || Ag^(+) |Ag` and
`therefore E_(cell)^(Theta) = E_("cathode")^(Theta) - E_("anode")^(Theta)`
`= 0.80 - (-0.13) = 0.93 V`