Question #bdd2f

May 17, 2015

Yes, lead metal can reduce the $A {g}^{2 +}$ cation, which in turn can oxidize the lead metal.

If you take a look in your standard reduction potential table, you'll notice that the redox potentials for these two reaction are

$A {g}_{\left(a q\right)}^{2 +} + 1 {e}^{-} r i g h t \le f t h a r p \infty n s A {g}_{\left(a q\right)}^{+}$, ${E}^{\circ} = \text{+1.98 V}$

$P {b}_{\left(a q\right)}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s P {b}_{\left(s\right)}$, ${E}^{\circ} = \text{-0.13 V}$

Now, regardless if you have a table that lists the most negative values of ${E}^{\circ}$ at the top or at the bottom, here's how you should look at this.

Notice that I've used equilibrium arrows for the two reactions.

In general, the more negative an electrode potential is, the more the equilibrium will lie to the left, meaning that the species that's on the right side of the equilibrium will lose electrons more easily.

Likewise, the more positive an electrode potential is, the more the equilibrium will lie to the right, meaning that the species that's on the right side will gain electrons more easily.

Now, look at your two reactions. The first reaction has a positive (or more positive, since we're not talking about absolute terms here) electrode potential, which means that $A {g}^{2 +}$ will gain an electron to form $A {g}^{+}$.

Likewise, the second reaction has a more negative electrode potential, which means that $P b$ will lose electrons to form $P {b}^{2 +}$.

When you put these two reactions together, you'll get a flow of electrons from $P {b}_{\left(s\right)}$ to $A {g}_{\left(a q\right)}^{2 +}$. In other words, lead metal will reduce $A {g}^{2 +}$ to $A {g}^{+}$.

At the same time, $A {g}^{2 +}$ will oxidize lead metal to $P {b}^{2 +}$.

May 17, 2015

I am assuming you mean $A {g}^{+}$ in which case the answer is yes and yes.

You should list the ${E}^{0}$ values from -ve to +ve. That way you will effectively get a numerical form of the activity series.

$P {b}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s P b$ ${E}^{0} = - 0.13 \text{V}$

$A {g}^{+} + e r i g h t \le f t h a r p \infty n s A g$ ${E}^{0} = + 0.8 \text{V}$

A useful rule is "bottom left will oxidise top right".

Which is the same as "top right will reduce bottom left".

You can see that Pb2+/Pb is the more -ve half reaction so will tend to go right to left and push out electrons.

Ag+/Ag is the more +ve half reaction so wil go left to right and take in electrons.

The overall reaction is therefore:

$P {b}_{\left(s\right)} + 2 A {g}_{\left(a q\right)}^{+} \rightarrow P {b}_{\left(a q\right)}^{2 +} + 2 A {g}_{\left(s\right)}$

This is the same as using the activity series where you would say that lead is above silver so will displace it from a solution of its salt.

The advantage of this method is that the arithmetic difference of the ${E}^{0}$ values gives us the e.m.f of the cell formed by combining the two 1/2 cells.