Use half-cell potentials to calculate #K_(eq)# for the reaction below?

#2Cu# (s)#+2H^+# (aq)#\toH_2# (g)

#E°_{Cu\toCu^(2+)}=-0.34# V


My work:
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2 Answers
Jul 29, 2018

Answer:

#K_(eq) = 2.16xx10^-12 #

Explanation:

#Cu^o(s) + 2H^+(aq) => Cu^(+2)(aq) + H_2(g)#

Using #E^o = (0.0592/n)logK_(eq)# => #log K_(eq) = ((n*E^o)/0.0592)#

For above reaction:
n = 2 and
#E^o(cell) = E^o("Reduction") - E^o("Oxidation")# = (0.00 volt) - (+0.34 volt) = -0.34 volt

#logK_(eq) = (2(-0.34))/0.0592 = -11.5#

#K_(eq) = 10^(-11.5) = 3.16xx10^-12#

Jul 30, 2018

Well, for this nonspontaneous reaction... #K = 3.16 xx 10^(-12)#.


First of all, let's re-balance the reaction, which should be:

#"Cu"(s) -> "Cu"^(2+)(aq) + 2e^(-)#, #E_(red)^@ = +"0.34 V"#
#ul(2"H"^(+)(aq) + 2e^(-) -> "H"_2(g))#, #E_(red)^@ = "0.00 V"#
#2"Cu"(s) + 2"H"^(+)(aq) -> 2"Cu"^(2+)(aq) + "H"_2(g)#

The reduction potential of copper cation is more positive, so copper would rather be reduced.

Hence, this reaction is nonspontaneous as-written if we try our hardest and certainly fail to oxidize copper using #"HCl"#.

The (nonspontaneous) cell potential is:

#E_(cell)^@#

#= overbrace(E_(red)^@)^"reduction" +overbrace(E_(o x)^@)^"oxidation"#, OR #overbrace(E_("cathode")^@)^"reduction" - overbrace(E_"anode"^@)^"reduction"#

#= "0.00 V" + (-"0.34 V")#, OR #"0.00 V" - (+"0.34 V")#

#= -"0.34 V"#

From this we obtain the positive #DeltaG_(rxn)^@#:

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

#= -(2 cancel("mol e"^(-)))/("1 mol Cu") cdot "96485 C/"cancel("mol e"^(-)) cdot (-"0.34 V")#

#=# #"65610 J/mol"#

As a result, since we are at equilibrium,

#cancel(DeltaG)^(0) = DeltaG^@ + RTlncancelQ^K#

and thus, we get the physically impractical equilibrium constant:

#color(blue)(K) = e^(-DeltaG^@//RT)#

#= e^(-(+"65610 J/mol")//("8.314 J/mol"cdot"K" cdot "298 K")#

#= e^(-26.5)#

#= color(blue)(3.16 xx 10^(-12))#