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.
