Question #e1b20

1 Answer
Jun 16, 2015

Answer:

Yes, you are spot on.

Explanation:

The net ionic equation for this redox reaction looks like this

#stackrel(color(blue)(0))(Zn_((s))) + stackrel(color(blue)(+2))(Pb_((aq))^(2+)) -> stackrel(color(blue)(+2))(Zn_((aq))^(2+)) + stackrel(color(blue)(0))(Pb_((s)))#

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(color(blue)(0))(Zn_((s))) -> stackrel(color(blue)(+2))(Zn_((aq))^(2+)) + 2e^(-)#, #" "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(color(blue)(+2))(Pb_((aq))^(2+)) + 2e^(-) -> stackrel(color(blue)(0))(Pb_((s)))# #" "E^0 = "-0.13 V"#

The cell potential will be

#E_"cell"^0 = E_"cathode" + E_"anode"#

#E_"cell"^0 = -"0.13 V" + "0.76 V" = "+0.63 V"#