# Question #7bc94

##### 2 Answers

All the voltages I've measured before in lab were positive, so the

Then, I would get

**DISCLAIMER:** *LONG ANSWER!*

Well, your

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 transferredper 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

When a reduction half-reaction is **reversed**, it becomes the corresponding **oxidation** half-reaction with

#=> 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

This means our

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

The full Nernst equation so far for

#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

#-(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

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

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

I get

#### Explanation:

There is an unfair catch to this question in that the cell diagram given is for the **non - spontaneous reaction** hence

This is wrong because

Look at the

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

The spontaneous cell reaction will be:

This means that the measured value of

Now we can use The Nernst Equation:

At 298K this simplifies to:

We can use the value of