A cell is constructed by dipping a zinc rod in 0.1 M zinc nitrate solution and a lead rod in 0.2 M lead nitrate solution.
`E_(Pb^(2+)//Pb)^(Theta) = -0.13 V and E_(Zn^(2+)//Zn)^(Theta) = -0.76V`
(i) Write the spontaneous cell reaction.
(ii) Calculate standard emf and emf of the cell.

Explanation

(i) `Zn^(2+) +2e^(-) to Zn, E^(Theta) = -0.76V`
`Pb^(2+) +2e^(-) to Pb, E^(Theta) = -0.13V`
From the standard electrode potentials, it is evident that the electrons will be released at the zinc electrode and gained at the lead electrode. Therefore, the spontaneous cell reaction is :
`Zn +Pb^(2+) to Zn^(2+) +Pb`
or cell representation is : `Zn//Zn^(2+)"||"Pb^(2+)//Pb`
(ii) `E_("cell")^(Theta)=E_("cathode")^(Theta)-E_("anode")^(Theta)`
`= -0.13V-(-0.76V)=0.63V`
`E_("cell")=E_("cell")^(Theta)-(0.0591)/(n)"log"(["Products"])/(["Reactants"])`
`=0.63-(0.0591)/(2)"log"([Zn^(2+)])/([Pb^(2+)])`
`=0.63 -(0.0591)/(2)"log"([0.1])/([0.2])`
`=0.63-0.0295 log ((1)/(2))V`
`=0.63-0.0295[log 1-log 2]V`
`=0.63 -0.0295 [0-0.3010]`
`=0.63+0.03=0.66V`