The electrochemical cell reaction of the Daniel cell is
`Zn_((s))+Cu^(2+)""_((aq)) rarr Zn^(2+)""_((aq))+Cu_((s))`
What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?

Explanation

`Zn_((s))+Cu^(2+)""_((aq)) rarr Zn^(2+)""_((aq))+Cu_((s))`
`E_("cell")=E^(@)-0.0591/nlog""([Zn^(2+)])/([Cu^(2+)])`
If the concentration of `Zn^(2+)` is increased, the second term on the RHS becomes more negative. This lowers the value of `E_(cell)`, so `E_(cell)` value will decrease if the ion concentration in the anode compartment is increased.