It depends on what ions you add and the half-cell to which you add them.
Heres an example that uses a Daniell cell such as that shown below.
The half-reactions are:
color(white)(mmmmmmmmmmmmmmmmmmmml)E^°"/V"
"Zn(s)" color(white)(mmmmmll)→ "Zn"^(2+)(aq) +2"e"^"-"; color(white)(ml)stackrel(————)(+0.76)
"Cu"^(2+)"(aq)" + 2"e"^"-"color(white)(ll) → "Cu(s)"; color(white)(mmmmmml)+0.34
stackrel(———————————————————-)("Zn(s)" + "Cu"^(2+)"(aq)" → "Zn"^(2+)"(aq)" + "Cu(s)"; +1.10)
Adding "Ag"^+ to the "Zn" half-cell
Now, let's add a solution of "AgNO"_3 to the "Zn" half-cell. We can get the following reaction:
color(white)(mmmmmmmmmmmmmmmmmmmmm)E^°"/V"
"Zn(s)"color(white)(mmmmmmll) → "Zn"^(2+)"(aq)" +"2e"^"-";color(white)(ml) stackrel(———)(+0.76)
"2Ag"^"+""(aq)" + "2e"^"-"color(white)(ml) → "2Ag(s)";color(white)(mmmmmll) +0.80
stackrel(————————————————————)("Zn"(s) + "2Ag"^"+""(aq)" → "Zn"²⁺"(aq)" + "2Ag(s)"; +1.56)
The zinc atoms will transfer their electrons directly to the silver ions, and solid "Ag" will plate out on the "Zn" electrode. The concentration of "Zn"^(2+) ions will increase.
According to Le Châtelier's Principle, this shifts the position of the "Zn-Cu" equilibrium to the left. The cell potential decreases.
Adding "Al"^(3+) to the "Zn" half-cell
If instead we add "Al"("NO"_3)_3 to the "Zn" half-cell, we get
color(white)(mmmmmmmmmmmmmmmmmmmmmmm)E°/V
#"3Zn"^(2+)"(aq)" + "6e"^"-"color(white)(ml) → "3Zn(s)";color(white)(mmmmmmmll) stackrelcolor(blue)(———)("-0.76")
"2Al(s)"color(white)(mmmmmmll) → "2Al"^(3+)"(aq)" + "6e"^"-"; color(white)(mll)+1.66
stackrel(————————————————————)("3Zn"^(2+)("aq") + "2Al(s)" → "3Zn(s)" + "2Al"^(3+)"(aq)"; +0.90)#
The "Al"^(3+) will react directly with the "Zn" electrode and form more "Zn"^(2+) ions.
Increasing ["Zn"^(2+)] will shift the position of the "Zn-Cu" equilibrium to the right.
The cell potential will increase.
