Question #b5579

1 Answer
Sep 6, 2015

Answer:

Here's what's going on here.

Explanation:

FULL QUESTION

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So, you are given four half-reactions and their associated standard reduction potentials, #E^@#.

For part (a) you need to figure out which combination of these four half-reaction will produce the largest cell potential. Now, you know that the standard cell potential can be calculated by using the reduction potential and oxidation potential of the two half-reactions.

All four half-reactions given to you are reduction half-reactions , which means that you're going to have to reverse them to get the oxidation half-reactions needed for the cell.

You know that for a half-reaction you have

#E_"oxidation"^@ = -E_"reduction"^@#

What this means is that you need to find the combination of half-reactions that gives the most positive value of

#E_"cell"^@ = E_"reduction"^@ + E_"oxidation"^@#

The most positive oxidation potential will belong to the reverse reaction that has the most negative reduction potential. Notice that the answer given to you for part (a) shows that the oxidation reaction is

#2 * ["Au"_text((s]) + 4"Br"_text((aq])^(-) -> "AuBr"_text(4(aq])^(-) + 3e^(-)]#

The oxidation potential is equal to

#E_"oxidation"^@ = -(-"0.858 V") = +"0.858 V"#

If you pair this oxidation half-reaction with the reduction half-reaction that has the most positive #E_"reduction"^@#, you will get the biggest value for #E_"cell"^@#.

The most positive reduction potential belongs to the third half-reaction

#3 * ["IO"_text(aq])^(-) + "H"_2"O"_text((l]) + 2e^(-) -> "I"_text((aq])^(-) + 2"OH"_text((aq])^(-)]#

which has #E_"reduction"^@ = +"0.49 V"#

This means that #E_"cell"^@# will be

#E_"cell"^@ = +"0.49 V" + "0.858 V" = "+1.35 V"#

Remember, for any redox reaction you have, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This is why the two half-reactions are multiplied by #2# and #3#, respectively. The total number of electrons transferred is equal to six.

For part (b), the exact same approach can be used. You need to find a combination of half-reactions that will produce the smallest positive cell potential.

Once again, you need to find your oxidation half-reaction by reversing one of the four reduction half-ractions. If you use the second reduction half-reaction as the oxidation half-reaction, you get

#2 * ["Eu"_text((aq])^(2+) -> "Eu"_text((aq])^(3+) + e^(-)]#

the oxidation potential will be

#E_"oxidation"^@ = -(-"0.43 V") = +"0.43V"#

The only possible option is to use the fourth reduction half-reaction

#"Sn"_text((aq])^(2+) + 2e^(-) -> "Sn"_text((s])#

which has

#E_"reduction"^@ = -"0.1364 V"#

The cell potential will be

#E_"cell"^@ = "0.43 V" + (-"0.1364 V") = +"0.29 V"#

It's very important to notice that you do not multiply the reduction or oxidation potentials when you multiply the half-reactions!

This is not like you would do with the enthalpy change of a reaction, for example, which changes accordingly if you multiply or divide the coefficients.

So, the idea is that a redox reaction needs a reduction half-reaction and an oxidation half-reaction. Something needs to lose electrons and something else needs to gain them.

Every time you have a reduction half-reaction and its associated reduction potential, keep in mind that you can use this reduction half-reaction as an oxidation half-reaction if you reverse it.

When you reverse a reduction half-reaction, you need to change the sign of its reduction potential, which now becomes its oxidation potential.

As far as for why they listed the anode first and the cathode second in the solution to part (a), and the cathode first and anode second in the solution to part (b), that's not really an important aspect.

As long as you keep track of which one is which, i.e. the reduction half-reaction and the oxidation half-reaction, the order in which you list them is not important.