# Question #e1b20

Jun 16, 2015

Yes, you are spot on.

#### Explanation:

The net ionic equation for this redox reaction looks like this

$\stackrel{\textcolor{b l u e}{0}}{Z {n}_{\left(s\right)}} + \stackrel{\textcolor{b l u e}{+ 2}}{P {b}_{\left(a q\right)}^{2 +}} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z {n}_{\left(a q\right)}^{2 +}} + \stackrel{\textcolor{b l u e}{0}}{P {b}_{\left(s\right)}}$

The more reactive zinc will remove the lead (II) ions from solution.

Notice that the oxidation state of zinc goes from zero on the reactants' side, to +2 on the products' side, which means that zinc is being oxidized.

The oxidation half-reaction, which takes place at the anode, will look like this

$\stackrel{\textcolor{b l u e}{0}}{Z {n}_{\left(s\right)}} \to \stackrel{\textcolor{b l u e}{+ 2}}{Z {n}_{\left(a q\right)}^{2 +}} + 2 {e}^{-}$, $\text{ "E^0 = "+0.76 V}$

On the other hand, lead's oxidation state goes from +2 on the reactants' side, to zero on the products' side, which means that it is being reduced.

The reduction half-reaction, which takes place at the cathode, will be

$\stackrel{\textcolor{b l u e}{+ 2}}{P {b}_{\left(a q\right)}^{2 +}} + 2 {e}^{-} \to \stackrel{\textcolor{b l u e}{0}}{P {b}_{\left(s\right)}}$ $\text{ "E^0 = "-0.13 V}$

The cell potential will be

${E}_{\text{cell"^0 = E_"cathode" + E_"anode}}$

${E}_{\text{cell"^0 = -"0.13 V" + "0.76 V" = "+0.63 V}}$