# Question 38b3d

Apr 18, 2014

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 ${\text{Ag}}^{+}$ to the $\text{Zn}$ half-cell

Now, let's add a solution of ${\text{AgNO}}_{3}$ to the $\text{Zn}$ 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 $\text{Ag}$ will plate out on the $\text{Zn}$ electrode. The concentration of ${\text{Zn}}^{2 +}$ ions will increase.

According to Le Châtelier's Principle, this shifts the position of the $\text{Zn-Cu}$ equilibrium to the left. The cell potential decreases.

Adding ${\text{Al}}^{3 +}$ to the $\text{Zn}$ half-cell

If instead we add "Al"("NO"_3)_3 to the $\text{Zn}$ 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 ${\text{Al}}^{3 +}$ will react directly with the $\text{Zn}$ electrode and form more ${\text{Zn}}^{2 +}$ ions.

Increasing $\left[{\text{Zn}}^{2 +}\right]$ will shift the position of the $\text{Zn-Cu}$ equilibrium to the right.

The cell potential will increase.