Question #e1b20

Question

Question #e1b20

1 Answer

Impact of this question

1947 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-16T08:33:30

Yes, you are spot on.

Explanation:

The net ionic equation for this redox reaction looks like this

$0 Z n_{(s)} + + 2 P b_{(a q)}^{2 +} \to + 2 Z n_{(a q)}^{2 +} + 0 P b_{(s)}$ $stackrel(color(blue)(0))(Zn_((s))) + stackrel(color(blue)(+2))(Pb_((aq))^(2+)) -> stackrel(color(blue)(+2))(Zn_((aq))^(2+)) + stackrel(color(blue)(0))(Pb_((s)))$

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

$0 Z n_{(s)} \to + 2 Z n_{(a q)}^{2 +} + 2 e^{-}$ $stackrel(color(blue)(0))(Zn_((s))) -> stackrel(color(blue)(+2))(Zn_((aq))^(2+)) + 2e^(-)$ , $E^{0} = +0.76 V$ $" "E^0 = "+0.76 V"$

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

$+ 2 P b_{(a q)}^{2 +} + 2 e^{-} \to 0 P b_{(s)}$ $stackrel(color(blue)(+2))(Pb_((aq))^(2+)) + 2e^(-) -> stackrel(color(blue)(0))(Pb_((s)))$ $E^{0} = -0.13 V$ $" "E^0 = "-0.13 V"$

The cell potential will be

$E_{cell}^{0} = E_{cathode} + E_{anode}$ $E_"cell"^0 = E_"cathode" + E_"anode"$

$E_{cell}^{0} = - 0.13 V + 0.76 V = +0.63 V$ $E_"cell"^0 = -"0.13 V" + "0.76 V" = "+0.63 V"$

Science

Math

Humanities

... and beyond

Question #e1b20

1 Answer

Explanation:

Related questions

Impact of this question