The electrochemical cell reaction of the Daniel cell is
`Zn(s) + Cu^(2+)(aq) to 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_((aq))^(2+) to Zn_((aq))^(2+) + Cu_((s))`
In this case `E_("cell")^(@) = 1.1 V`
Reaction quotient Q for the above reaction is , `Q = ([Zn^(2+)])/([Cu^(2+)])`
`:. E_("cell") = E_("cell")^@ - (0.0591)/(n) log ([Zn^(2+)])/([Cu^(2+)])`
If Suppose concentration of `Cu^(2+)` is 1.0 M then the concentration of `Zn^(2+)` is 10M (why because, ion concentration in the anode compartment increased by a 10 factor)
`:. E_("cell") = 1.1 - (0.0591)/2 log (10/1)`
`= 1.1 - 0.02955 (1)`
`= 1.070 V` (cell voltage decreased)
Thus, the initial volatage is greater than `E^@` because `Q < l`. As the reaction proceeds, `[Zn^(2+)]` in the anode compartment increases as the zinc electrode dissovles, while `[Cu^(2+)]` in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the `Q = [Zn^(2+)][Cu^(2+)]` steadily increases and the cell voltage therefore steadily decreases.
`[Zn^(2+)]` will continue to increase in the anode compartment and `[Cu^(2+)]` will continue to decrease in the cathode compartment. Thus the value of Q will increase further leading to a further decrease in `E_("cell")` value.