FRQ - Electrochemistry?

a. Which occurs more often in nature, oxidation or reduction? Discuss the answer.
b. The half-reaction for the hydrogen electrode, 2#H^+ + 2e^-##->##H_2#, has been given the #E^@#value of zero. Explain why such a standard is necessary.
c. They system #Mg(s)+Cu^(2+) -> Mg^2+ (aq)+ Cu(s)# has a value #E^@cell# of + 2.71V. What is the value of #DeltaG^@# for his system?

1 Answer
Truong-Son N.
Apr 20, 2018

Well, how much oxygen do we have in the air? What does oxygen do to metals?

From that, you can answer #(a)# on your own time.

#b)# We measure potential differences. Differences are always arbitrary and relative. If we have no reference state, then the measurement is meaningless. Having #E_("SHE")^@ = "0 V"# is of course convenient, because #"anything" + 0 = "itself"#.

#c)# We know that at #25^@ "C"# for

#"Mg"(s) + "Cu"^(2+)(aq) -> "Cu"(s) + "Mg"^(2+)(aq)#,

we have

#DeltaG^@ = -nFE_(cell)^@#,

where #n# is the mols of electrons transferred per mol of atoms. #F# is the Faraday constant.

We can see that #2# electrons are transferred. Magnesium gets oxidized by transferring electrons away, and into #"Cu"^(2+)# to reduce it. That makes #"Cu"^(2+)# the oxidizing agent, and you should be able to then identify the reducing agent.

#"Mg"(s) -> "Mg"^(2+)(aq) + 2e^(-)#
#ul("Cu"^(2+)(aq) + 2e^(-)(aq) -> "Cu"(s))#
#"Mg"(s) + "Cu"^(2+)(aq) -> "Cu"(s) + "Mg"^(2+)(aq)#

Therefore,

#DeltaG^@ = -("2 mol e"^(-)"/mol Cu")("96485 C/mol e"^(-))("2.71 V")#

#= ?#

Consider what units you get here (show that you get #"J/mol"# by noting that #"J" = "V"cdot"C"#), and then you should convert into #"kJ/mol"# by convention.

This value will be negative, because the combination favors the forward reaction at room temperature. What does that tell you about what #"Mg"# prefers? Does it want to get oxidized or reduced more than #"Cu"#?

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