Question

Question #7bc94

Answer 1

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"`.

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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")`

Answer 2

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")`