Question

A zinc rod is placed in 0.1 MZnSO4 solution at 298 K. Write the electrode reaction and calculate the potential of the electrode. (EZ24/2 = -0.76V.)

Answer 1

Answer: The potential of electrode is -0.79 V

Step-by-step explanation:

When zinc is dipped in zinc sulfate solution, the electrode formed is
`Zn^(2+)(aq.)/Zn(s)`

Reduction reaction follows:
`Zn^(2+)(0.1M)+2e^-\rightarrow Zn(s);(E^o_(Zn^(2+)/Zn)=-0.76V)`

To calculate the potential of electrode, we use the equation given by Nernst equation:

`E_((Zn^(2+)/Zn))=E^o_((Zn^(2+)/Zn))-(0.059)/(n)\log ([Zn])/([Zn^(2+)])`

where,

`E_(cell)` = electrode potential of the cell = ?V

`E^o_(cell)` = standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

`[Zn]=1M` (concentration of pure solids are taken as 1)

`[Zn^(2+)]=0.1M`

Putting values in above equation, we get:

`E_(cell)=-0.76-(0.059)/(2)* \log((1)/(0.1))\\\\E_(cell)=-0.79V`

Hence, the potential of electrode is -0.79 V