Question #7bc94

2 Answers
Jun 7, 2017

All the voltages I've measured before in lab were positive, so the #E_(cell)# given to you doesn't make physical sense to use as-is, and you should flip the sign.

Then, I would get #"0.19 M"#.


DISCLAIMER: LONG ANSWER!

Well, your #"H"_2(g)# is listed as #"0.807 atm"#, which is a pressure... but that's OK, because #ln# does not work on arguments with units, and so each concentration is divided by a standard value, such as #c/"1 M"# or #P/"1 atm"#.

Also, you can't square root a negative number...

The Nernst equation relates the standard cell potential to the nonstandard cell potential. In general, it is:

#E_(cell) = E_(cell)^@ - (RT)/(nF)lnQ#,

where:

  • #E_(cell)# is the cell potential in nonstandard conditions.
  • #E_(cell)^@# is the standard cell potential, for #"1 M"# concentrations, #"1 atm"#, and #25^@ "C"#.
  • #R = "8.314472 J/mol"cdot"K"# is the universal gas constant.
  • #T# is the temperature in #"K"#.
  • #n# is the mols of electrons transferred per half-reaction (no sign).
  • #F = "96485 C/mol e"^(-)# is Faraday's constant.
  • #Q# is the reaction quotient, i.e. the not-yet-equilibrium constant.

What you'll need to do first is write your full reaction. The half-reactions were given:

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

#"Cd"^(2+)(aq) + 2e^(-) -> "Cd"(s)#, #E_(red)^@ = -"0.403 V"#

For a spontaneous reaction, since #DeltaG^@ = -nFE_(cell)^@# at #25^@ "C"# and #"1 atm"#, and #DeltaG^@ < 0# indicates spontaneity at #25^@ "C"# and #"1 atm"#, #E_(cell)^@# must be positive at #25^@ "C"# and #"1 atm"#.

When a reduction half-reaction is reversed, it becomes the corresponding oxidation half-reaction with #E_(o x)^@ = -E_(red)^@#.

#=> E_"cell"^@ = E_(red)^@ + E_(o x)^@#

#= "0.00 V" + (-(-"0.403 V")) = +"0.403 V"#

Upon reversing the cadmium half-reaction, we now have the full reaction:

#2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g)#
#"Cd"(s) -> "Cd"^(2+)(aq) + cancel(2e^(-))#
#"----------------------------------------"#
#"Cd"(s) + 2"H"^(+)(aq) -> "Cd"^(2+)(aq) + "H"_2(g)#

So, its reaction quotient is:

#Q_c = ((["Cd"^(2+)]"/""1 M") (P_(H_2)"/"P^@))/((["H"^(+)]"/""1 M")^2)#

Your #Q#, however, was upside-down. It should be products over reactants.

This means our #Q# is:

#Q = ((1.00)(0.807))/(["H"^(+)]^2)#

The full Nernst equation so far for #"1 mol"# of #"Cd"(s)# is:

#E_(cell) = color(red)(+)"0.3631 V"# (given, corrected)

#= overbrace"0.403 V"^(E_(cell)^@) - ("8.314472 J/"overbrace(cancel"mol")^"accounted for"cdotcancel"K"cdot298.15 cancel"K")/(2 cancel("mol e"^(-))cdot"96485 C/"cancel("mol e"^(-)))ln((("1.00")("0.807"))/(["H"^(+)]^2))#

#=# #overbrace"0.403 V"^(E_(cell)^@) - "0.01285 V"ln((("1.00")("0.807"))/(["H"^(+)]^2))#

Solving for #["H"^(+)]#:

#-(E_(cell) - E_(cell)^@)#

#= -(+"0.3631 V" - "0.403 V") = +"0.0399 V"#

#= + "0.01285 V"ln((("1.00 M")("0.03299 M"))/(["H"^(+)]^2))#

#=> 3.105 = ln((("1.00 M")("0.03299 M"))/(["H"^(+)]^2))#

#=> e^(3.105) = (("1.00 M")("0.03299 M"))/(["H"^(+)]^2)#

This gives us an #"H"^(+)# concentration of...

#=> color(blue)(["H"^(+)]) = sqrt((("1.00")("0.807"))/e^(3.105)) " M"#

#=# #color(blue)("0.19 M")#

Jun 8, 2017

Answer:

I get #sf([H^+]=0.19color(white)(x)"mol/l")#

Explanation:

There is an unfair catch to this question in that the cell diagram given is for the non - spontaneous reaction hence #sf(E_(cell)^@)# is -ve.

This is wrong because #sf(E_(cell))# is an empirically measured quantity and must always have a +ve value. If you were to construct this cell in the laboratory and obtained a -ve value for its emf it means you have connected your voltmeter the wrong way round (AF).

Look at the #sf(E^@)# values:

#sf(Cd^(2+)+2erightleftharpoonsCd" "E^@=-0.403color(white)(x)V)#

#sf(2H^(+)+2erightleftharpoonsH_2" "E^@=" "0.00color(white)(x)V)#

These values indicate that the Cd 1/2 cell is the more -ve so will tend to push out electrons and shift right to left.

The H 1/2 cell will take in these electrons and shift left to right. This is why I do not use the term "reduction potentials" and I use the #sf(rightleftharpoons)# rather than an arrow to signify they can go in either direction, depending on what they are coupled with.

#sf(E_(cell)^@)# is the arithmetic difference between the 2 standard electrode potentials:

#sf(E_(cell)^@=0.00-(-0.403)=+0.403color(white)(x)V)#

The spontaneous cell reaction will be:

#sf(Cd+2H^+rarrCd^(2+)+H_2)#

This means that the measured value of #sf(E_(cell)=+0.3631color(white)(x)V)#

Now we can use The Nernst Equation:

#sf(E_(cell)=E_(cell)^@-(RT)/(zF)lnQ)#

At 298K this simplifies to:

#sf(E_(cell)=E_(cell)^@-0.0591/(z)logQ)#

#:.##sf(0.3631=0.403-0.0591/(2)logQ)#

#sf(logQ=(2xx0.0399)/(0.0591)=1.35)#

#:.##sf(Q=22.4)#

#sf(Q=([Cd^(2+)]xxpH_2)/([H^+]^2)=22.4)#

We can use the value of #sf(pH_2)# given since this is normalised against 1 atmosphere so is dimensionless.

#sf([H^+]^2=(1.00xx0.807)/(22.4)=0.03602)#

#sf([H^+]=sqrt(0.03602)=0.19color(white)(x)"mol/l")#