Question

What will be the cell potential for a cu−zn galvanic cell with [zn2+]=0.1 m and the [cu2+]=0.01 m? view available hint(s)?

Answer 1

Answer: 1.07 V

Explanation:
`E^0_([Zn^(2+)/Zn)=-0.76V`

`E^0_([Cu^(2+)/Cu])=+0.34V`

The metal with negative reduction potential will easily lose electrons and thus is oxidized and the one with positive reduction potential will easily gain electrons and thus is reduced.

`Zn+Cu^(2+)\rightarrow Zn^(2+)+Cu`

`E^o_(cell)` = standard electrode potential =
`E^0_(cathode)- E^0_(anode)=0.34-(-0.76)=1.1V`

Using Nernst equation:

`E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Cu^(2+)])`

where,

n = number of electrons in oxidation-reduction reaction = 2

`E_(cell)=1.10-(0.0592)/(2)\log ([0.1])/([0.01])`

`E_(cell)=1.07V`