Electrochemistry question? (Voltaic Cell)

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A: Which electrode is the cathode in the reaction above?

B: Give the half reaction occurring at each of the electrodes below. Make sure it shows the direction the reaction actually proceeds in the cell above.

C: Write a balanced chemical equation describing the total reaction associated with the electrochemical cell.

D: What is the EMF of the cell above if it is operated under standard state conditions?

1 Answer
Apr 10, 2017

I will break it down for you so you can understand how to look at these types of questions.

Explanation:

This is a Galvanic cell. Galvanic cells are spontaneous meaning they have a -DeltaG^@. A -DeltaG^@ just means the reaction occurs without the input of any outside energy source, under standard conditions that is. The standard cell potential of the cell, E^@, is always positive in a galvanic cell. Basically, Gibbs and Standard Cell Potential have opposite signs.

You have solid "Cu" and "Zn" in their respective half cells. Looking at the standard reduction potentials provided, the more positive value means the more likely that species is going to be reduced. A quick look at the table will show you that "Cu"^(+2) ions will readily reduce to form solid "Cu" than "Zn"^(+2) ions will to become "Zn" solid.

color(white)(aaaa)color(magenta)["Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s) color(white)(aaaaa)"E"^@("red") = +0.34] ("more positive red. potential means reduction will occur")

color(white)(aaaa)color(green)["Zn"^(+2)(aq) + 2e^(-)->"Zn"(s) color(white)(aaaaa)"E"^@("red") = -0.76] ("less positive red. potential means oxidation occurs")

color(white)(aaaaaaaaaaaaaaaa)
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color(blue)("A: Which electrode is the cathode in the reaction above?")

  • Since we know that"Cu"^(+2) ions are more likely to be reduced to become "Cu"(s), the color(red)("Cu electrode is labeled as the cathode") while the color(red)("Zn electrode is labeled as the anode"), where oxidation occurs. (remember the memory aid AN OX, RED CAT)

color(blue)("B: Give the half reaction occurring at each of the electrodes below.")

color(blue)("Make sure it shows the direction the reaction actually proceeds in the")
color(blue)("cell above.")

  • If reduction is occurring at the "Cu" cathode (where we know the solution of "Cu(NO3)"_2 ionizes to give off "Cu"^(+2) cations and "NO"_3^(-1) anions), this means "Cu"^(+2) ions will gain 2e^- and will deposit onto the "Cu" electrode. The half reaction is the following for the left half cell:

color(white)(aaaaaaaaaaaa)color(red)["Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s)

color(white)(aaaaaaaaaaaa)

  • In the right half cell, the "Zn" electrode (which is the anode) will lose 2e^- and "Zn"^(+2) ions will fall out into the solution which already has "Zn"^(+2) cations and "NO"_3^(-1) ions from the zinc nitrate solution. The half reaction in the right cell is the following:

color(white)(aaaaaaaaaaa)color(red)["Zn"(s)-> "Zn"^(+2)(aq) + 2e^(-)

color(blue)("C: Write a balanced chemical equation describing the total reaction of the")
color(blue)("electrochemical cell.")

  • If we combine the two half reactions, we get

"Cu"^(+2)(aq) + cancel(2e^(-)) ->"Cu"(s)
"Zn"(s)-> "Zn"^(+2)(aq) + cancel(2e^(-))
--------------
color(red)["Cu"^(+2)(aq) + "Zn"(s)->"Cu"(s) + "Zn"^(+2)(aq)

color(blue)("D: What is the EMF of the cell above(under standard state conditions)")

  • Okay. We have established that at the "Cu" half cell, reduction will occur and at the "Zn" half cell. oxidation will occur. To find the EMF of the cell, the following equation is going to be used.

color(white)(aaaaaaa)"E"^@("cell") = "E"^@("red") + "E"^@("oxidation")

Since we were provided "E"^@("red") potentials, we have to reverse the reaction to show the oxidation of "Zn"(s) as well as reversing the value of the standard reduction potential to get the standard oxidation potential.

"Cu"^(+2)(aq) + 2e^(-) ->"Cu"(s)color(white)(aaaaa)"E"^@("red") = +0.34

"Zn"(s)->"Zn"^(+2)(aq) + 2e^(-)color(white)(aaaaa)"E"^@("oxidation") = +0.76
color(white)(aaaaa)
- Use the equation to plug in and solve:

color(white)(aaaaa)"E"^@("cell") = "E"^@("red") + "E"^@("oxidation")

color(white)(aaaaa)color(red)["E"^@("cell") = (+0.34) + (+0.76) = +1.1