source : socratic.org
Electrochemistry question? (Voltaic Cell) | Socratic
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(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)#
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#
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