Question #d9281
1 Answer
Explanation:
The idea here is that you need to use the standard reduction potentials,
You know that you have
#"Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Zn"_ ((s))" "E^@ = -"0.76 V"#
#"Pb"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V"#
#"Cu"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V"#
Now, the standard reduction potentials measure the tendency of a chemical species to release electrons and form cations when compared to that of hydrogen.
A negative
A positive
Now, chemical species that lose electrons more readily are stronger reducing agents than chemical species that tend to lose electrons less readily.
When you compare the
- the equilibrium with the less positive / more negative
#E^@# value will lie further to the left- the equilibrium with the less negative / more positive
#E^@# value will lie further to the right
Let's take the first two reduction equilibria. You have
#E^@("Zn"^(2+), "Zn") = -"0.76 V"#
#E^@("Pb"^(2+), "Pb") = -"0.13 V"#
Here
You can thus say that zinc metal,
Consequently, zinc will also reduce
In other words, when zinc metal is placed in a solution that contains
This implies that zinc is metal
Finally, notice that metal
#"Metal A "-> " Lead"#
#"Metal B " -> " Zinc"#
#"Metal C " -> " Copper"#
So, to answer such problems quickly, list the reduction half-reactions in order of increasing
#"Zn"_ ((aq))^(2+) + 2"e"^(-) rightleftharpoons color(blue)("Zn"_ ((s)))" "E^@ = -"0.76 V"#
#color(red)("Pb"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Pb"_ ((s))" "E^@ = -"0.13 V"#
#color(red)("Cu"_ ((aq))^(2+)) + 2"e"^(-) rightleftharpoons "Cu"_ ((s))" "E^@ = +"0.34 V"#
and keep in mind that the chemical species listed in the top right can reduce the chemical species listed in the bottom left.
For example, zinc metal is
Similarly, you can say that the chemical species located in the bottom left can oxidize the chemical species located to the top right.
Here
#{(color(red)("Bottom left")color(white)(a)"oxidizes" color(white)(a)color(blue)("top right ")), (color(blue)("Top right")color(white)(a)"reduces" color(white)(a)color(red)("bottom left")) :}#