p 118-120


1.         a.         reducing agents are themselves oxidized. From the direction of the reaction, it is Cu that gets oxidized, not Fe+2. So, yes Cu is a stronger reducing agent than Fe+2.

            b.         Most of the time neutral metals are better reducing agents. But not with gold! From the direction of the reaction, it is  Fe+2 that gets oxidized, not Au. So, yes Fe+2  is a stronger reducing agent than Au.

            c.         The reverse reaction is favoured. That should be obvious from what we argued in (a) and (b). If it's not obvious, you are being distracted by the Bruins' domination of the Habs. You can also check the E values.

            d.         The strongest oxidizing agent is the one that's best at removing electrons. So the answer is Au+3.


            Another way of convincing yourself is : Au+3 + 3e --> Au      E = +1.50 V. This E value is bigger than anything else from your table except fluorine. So from your table, only fluorine is a better oxidizing agent than Au+3.


2.         a.         Li

            b.         Li. If Cs was stronger the reverse reaction would occur. Also, Li--> Li+1 + 1e  ( E = 3.04 V: the largest value)

            c.         3Li + Al+3 --> 3 Li+1 + Al

            d.         Al+3 is hungrier for electrons than Li+1 is.


3..        When they tell you that A + D+2 react, then A + D+2-->A+2+D. This implies that D+2is better at getting reduced than A+2. If you apply this to every reaction, you realize that D+2>B+2>E+2>C+2

4.         Cr + Al+3 à no reaction, so Al is better than Cr at losing electrons

            Mg + Al+3 à reaction, so Mg is better than Al at losing electrons

            Ni+2 + Al à reaction, so Al is better than Ni at losing electrons

            Ni + Cr+3 à no reaction, so Cr is better than Ni at losing electrons

            So from the best to the worst at losing electrons( best reducing agent) we have

            Mg> Al>Cr> Ni

5.         To answer the question you need to see from the table of reduction potentials that Cu+2 is a better oxidizing agent than Cr+3. So if you are wondering whether you should store a solution containing Cr+3 ( from Cr(NO3)3))in copper, you're really wondering whether the following will happen:
Cr+3 + Cu --> Cu+2 + Cr

But it won't because Cu+2 wins out. So the answer is that, yes, the     scientist should store the solution in the copper container, since nothing         will happen to it.

6.         Ag+ + 1e à Ag           E = 0.80 V

            Zn à Zn+2 + 2e           E = 0.76 V


            produce a total of 1.56 V, greater than any other combination. Unfortunately they did not list it as one of the choices. So the answer is B


7.         Use      Cr+2 + 2e à Cr           E = -0.91 V

                        Pb+2 + 2e à Pb           E = -0.13 V,


            So you need to reverse the first one to obtain a positive overall E value.

            Cr à   Cr+2 + 2e                     E = +0.91 V

            Pb+2 + 2e à Pb                       E = -0.13 V


            Overall:           Cr + Pb+2 à Cr+2 + Pb            E = 0.78 V


            1.         anode: Cr à   Cr+2 + 2e

            2.         Pb

            3.         Pb+2

                    4.         Cr + Pb+2 à Cr+2 + Pb

            5.         E = 0.78 V