A galvanic cell consists of a metallic zinc plate immersed in 0.1 M `Zn (NO_(3))_(2)` solution and metallic plate of lead in 0.02M `Pb(NO_(3))_(2)` solution. Calculate the emf of the cell.
Write the chemical equation for the electrode reactions and represent the cell.
`("Given" : E^(@) Zn^(2+)ZN=0.76V, " " E^(@)Pb^(2+)//Pb = -0.13V)`

Answer

0.61`" V "`

Explanation

Cell reaction : `Zn(s)+Pb^(2+)(0.02M)to Zn^(2+)(0.1" M")+Pb(s)`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Pb^(2+)])=[-0.13-(0.76)]-(0.0591)/(2)"log"(0.1)/(0.02)`
`=0.63-0.02955" log "5=0.63-0.02955xx0.69877`
`=0.63-0.02065=0.609345" V "=0.61" V "`.