Question

The standard electrode potential of electrode

\[\ce{\underset{(0.02M)}{Zn^2+ (aq)} || Zn(s)}\] is -0.76 V. Calculate its electrode potential.

Answer 1

\[\ce{Zn++(aq) + 2e- -> Zn(s); E^0_{Zn}= -0.76 V}\]

Here, [Zn] = 1, Number of electrons = n = 2

According to Nernst equation,

`E_(Zn)=E_(Zn)^0 - 0.0592/n log_10  ([Zn])/([Zn^(2+)])`

`= - 0.76-0.0592/2log_10  1/0.02`

`=- 0.76-0.0592/2log_10 (50)`

`= - 0.76 - 0.0503`

`E_(Zn) = -0.8103 V`