Question #0aadc

The first thing to do when dealing with similar problems is to write out the reduction equilibria given to you

#"Ce"_ ((aq))^(4+) + color(white)(1)"e"^(-) rightleftharpoons "Ce"_ ((aq))^(3+)" "E^@ = +"1.72 V"#

#"Fe"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Fe"_ ((s))" "E^@ = -"0.44 V"#

For a given reduction equilibrium, the value of the standard reduction potential, #E^@#, tells you the position of the equilibrium in relation to a reference hydrogen electrode.

When #E^@# is positive, the equilibrium lies to the right, meaning that the chemical species loses electrons less readily than hydrogen.

When #E^@# is negative, the equilibrium lies to the left, meaning that the chemical species loses electrons more readily than hydrogen.

Now, when you compare two #E^@# values, you should know that

• the reduction equilibrium that has the less negative / more positive #E^@# value will shift to the right
• the reduction equilibrium that the more negative / less positive #E^@# value will shift to the left

In your case, the #E^@# value for the first equilibrium is negative and the #E^@# value for the second equilibrium is positive. This tells you that when you connect these two half-cells, you will get

#"Ce"_ ((aq))^(4+) + color(white)(1)"e"^(-) rightleftharpoons "Ce"_ ((aq))^(3+)" "E_"red"^@ = +"1.72 V"#

#stackrel( color(red)( rarr ))(color(white)(aacolor(darkgreen)("shift to the right")aaaa))#

Since the forward reaction proceeds here, you can say that this is your reduction half-reaction

#"Ce"_ ((aq))^(3+) + color(white)(1)"e"^(-) -> "Ce"_ ((aq))^(3+)" "E_"red"^@ = +"1.72 V"#

Likewise, you will get

#"Fe"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Fe"_ ((s))" "E_"oxi"^@ = -(-"0.44 V") = +"0.44 V"#

#stackrel( color(red)( larr ))(color(white)(aacolor(darkgreen)("shift to the left")aaaa))#

The reverse reaction proceeds here, so reverse the sign of #E^@# and write the oxidation half-reaction

#"Fe"_ ((s)) -> "Fe"_ ((aq))^(2+) + 2"e"^(-)" "E_"oxi"^@ = +"0.44 V"#

You can now balance and add the two half-reactions to get

#{("Ce"_ ((aq))^(4+) + color(white)(1)"e"^(-) -> "Ce"_ ((aq))^(3+)" " " " " "| xx 2), (color(white)(aaaaaa)"Fe"_ ((s)) -> "Fe"_ ((aq))^(2+) + 2"e"^(-)):}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#2"Ce"_ ((aq))^(4+) + "Fe"_ ((s)) + color(red)(cancel(color(black)(2"e"^(-)))) -> 2"Ce"_ ((aq))^(3+) + "Fe"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-))))#

You thus have

#2"Ce"_ ((aq))^(4+) + "Fe"_ ((s)) -> 2"Ce"_ ((aq))^(3+) + "Fe"_ ((aq))^(2+) #

The standard cell potential will be equal to

#E_"cell"^@ = E_"red"^@ + E_"oxi"^@#

#E_"cell"^@ = +"1.72 V" + "0.44 V" = "2.16 V#

Now you're ready to answer the questions given to you.

• #"Ce"^(4+)# is a weaker oxidizing agent than #"Fe"^(2+)#

This one is not true. The fact that the reaction proceeds in this direction, i.e. #E_"cell"^@ > 0#, which implies that this redox reaction is spontaneuos, tells you that #"Ce"^(4+)# is a stronger oxidizing agent than #"Fe"^(2+)#.

Similarly, iron metal, #"Fe"#, is a stronger reducing agent than #"Ce"^(3+)#. Kepp in mind that for a spontaneous redox reaction, you have

#overbrace(2"Ce"_ ((aq))^(4+))^(color(blue)("stronger oxi agent")) + underbrace("Fe"_ ((s)))_ (color(purple)("stronger red agent")) -> underbrace(2"Ce"_ ((aq))^(3+))_ (color(purple)("weaker red agent")) + overbrace("Fe"_ ((aq))^(2+))^(color(blue)("weaker oxi agent"))#

• #"Ce"^(4+)# will reduce #"Fe"^(2+)#

This one is also not true. #"Ce"^(4+)# cannot reduce #"Fe"^(2+)# because it oxidizes #"Fe"# to #"Fe"^(2+)#.

• #"Ce"^(4+)# is a stronger oxidizng agent than #"Fe"^(2+)#

This one is true, as shown above.

• #"Ce"^(4+)# will oxidize #"Fe"#

This one is also true, as shown above.