Question 7bc94

Jun 7, 2017

All the voltages I've measured before in lab were positive, so the ${E}_{c e l l}$ given to you doesn't make physical sense to use as-is, and you should flip the sign.

Then, I would get $\text{0.19 M}$.

Well, your ${\text{H}}_{2} \left(g\right)$ is listed as $\text{0.807 atm}$, 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 $\frac{c}{\text{1 M}}$ or $\frac{P}{\text{1 atm}}$.

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:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$,

where:

• ${E}_{c e l l}$ is the cell potential in nonstandard conditions.
• ${E}_{c e l l}^{\circ}$ is the standard cell potential, for $\text{1 M}$ concentrations, $\text{1 atm}$, and ${25}^{\circ} \text{C}$.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.
• $T$ is the temperature in $\text{K}$.
• $n$ is the mols of electrons transferred per half-reaction (no sign).
• $F = {\text{96485 C/mol e}}^{-}$ 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:

$2 {\text{H"^(+)(aq) + 2e^(-) -> "H}}_{2} \left(g\right)$, ${E}_{red}^{\circ} = \text{0.00 V}$

$\text{Cd"^(2+)(aq) + 2e^(-) -> "Cd} \left(s\right)$, ${E}_{red}^{\circ} = - \text{0.403 V}$

For a spontaneous reaction, since $\Delta {G}^{\circ} = - n F {E}_{c e l l}^{\circ}$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, and $\Delta {G}^{\circ} < 0$ indicates spontaneity at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, ${E}_{c e l l}^{\circ}$ must be positive at ${25}^{\circ} \text{C}$ and $\text{1 atm}$.

When a reduction half-reaction is reversed, it becomes the corresponding oxidation half-reaction with ${E}_{o x}^{\circ} = - {E}_{red}^{\circ}$.

$\implies {E}_{\text{cell}}^{\circ} = {E}_{red}^{\circ} + {E}_{o x}^{\circ}$

$= \text{0.00 V" + (-(-"0.403 V")) = +"0.403 V}$

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

$2 {\text{H"^(+)(aq) + cancel(2e^(-)) -> "H}}_{2} \left(g\right)$
${\text{Cd"(s) -> "Cd}}^{2 +} \left(a q\right) + \cancel{2 {e}^{-}}$
$\text{----------------------------------------}$
${\text{Cd"(s) + 2"H"^(+)(aq) -> "Cd"^(2+)(aq) + "H}}_{2} \left(g\right)$

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:

$Q = \frac{\left(1.00\right) \left(0.807\right)}{{\left[{\text{H}}^{+}\right]}^{2}}$

The full Nernst equation so far for $\text{1 mol}$ of $\text{Cd} \left(s\right)$ is:

${E}_{c e l l} = \textcolor{red}{+} \text{0.3631 V}$ (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 $\left[{\text{H}}^{+}\right]$:

$- \left({E}_{c e l l} - {E}_{c e l l}^{\circ}\right)$

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

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

$\implies 3.105 = \ln \left(\left({\left({\text{1.00 M")("0.03299 M"))/(["H}}^{+}\right]}^{2}\right)\right)$

$\implies {e}^{3.105} = \left({\left({\text{1.00 M")("0.03299 M"))/(["H}}^{+}\right]}^{2}\right)$

This gives us an ${\text{H}}^{+}$ concentration of...

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

$=$ $\textcolor{b l u e}{\text{0.19 M}}$

Jun 8, 2017

I get $\textsf{\left[{H}^{+}\right] = 0.19 \textcolor{w h i t e}{x} \text{mol/l}}$

Explanation:

There is an unfair catch to this question in that the cell diagram given is for the non - spontaneous reaction hence $\textsf{{E}_{c e l l}^{\circ}}$ is -ve.

This is wrong because $\textsf{{E}_{c e l l}}$ 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 $\textsf{{E}^{\circ}}$ values:

$\textsf{C {d}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C d \text{ } {E}^{\circ} = - 0.403 \textcolor{w h i t e}{x} V}$

$\textsf{2 {H}^{+} + 2 e r i g h t \le f t h a r p \infty n s {H}_{2} \text{ "E^@=" } 0.00 \textcolor{w h i t e}{x} V}$

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 $\textsf{r i g h t \le f t h a r p \infty n s}$ rather than an arrow to signify they can go in either direction, depending on what they are coupled with.

$\textsf{{E}_{c e l l}^{\circ}}$ is the arithmetic difference between the 2 standard electrode potentials:

$\textsf{{E}_{c e l l}^{\circ} = 0.00 - \left(- 0.403\right) = + 0.403 \textcolor{w h i t e}{x} V}$

The spontaneous cell reaction will be:

$\textsf{C d + 2 {H}^{+} \rightarrow C {d}^{2 +} + {H}_{2}}$

This means that the measured value of $\textsf{{E}_{c e l l} = + 0.3631 \textcolor{w h i t e}{x} V}$

Now we can use The Nernst Equation:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{z F} \ln Q}$

At 298K this simplifies to:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.0591}{z} \log Q}$

$\therefore$$\textsf{0.3631 = 0.403 - \frac{0.0591}{2} \log Q}$

$\textsf{\log Q = \frac{2 \times 0.0399}{0.0591} = 1.35}$

$\therefore$$\textsf{Q = 22.4}$

$\textsf{Q = \frac{\left[C {d}^{2 +}\right] \times p {H}_{2}}{{\left[{H}^{+}\right]}^{2}} = 22.4}$

We can use the value of $\textsf{p {H}_{2}}$ given since this is normalised against 1 atmosphere so is dimensionless.

$\textsf{{\left[{H}^{+}\right]}^{2} = \frac{1.00 \times 0.807}{22.4} = 0.03602}$

$\textsf{\left[{H}^{+}\right] = \sqrt{0.03602} = 0.19 \textcolor{w h i t e}{x} \text{mol/l}}$