The voltage generated by the zinc concentration cell described by,

Zn(s)|Zn2+ (aq, 0.100 M)||Zn2+ (aq, ____ M)|Zn(s)
is 24.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

Hi, I posted this question awhile back but I think it got drowned out so I didn't want to comment on it again for fear that it'd go unseen.
DrBob told me the value should be above 0.1 M.. I tried again and I was wondering if this is correct:

0.024 = (8.314*298 K)/(2*96485)
0.024 = 0.012839 * ln[x/0.1]
x = 0.648 M?

thanks ahead of time!

That answer is right

Well, well, looks like you're trying to zap me with some electric chemistry! Don't worry, I'll humor you with an answer.

First, let's take a look at your calculations. You're using the Nernst equation, trying to solve for the concentration of Zn2+ at the cathode. Quite electrifying, I must say!

Now, let's break it down. The Nernst equation is given by:

E = E° - (RT/nF) * ln(Q)

Where:
E is the cell potential (in volts)
E° is the standard cell potential (in volts)
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature (in Kelvin)
n is the number of electrons transferred in the reaction
F is Faraday's constant (96485 C/mol)
Q is the reaction quotient

In your equation, you correctly substituted the values, but there's a slight mix-up. You've assumed the reaction quotient (Q) to be the concentration of Zn2+ at the cathode (x) divided by the concentration of Zn2+ at the anode (0.1 M). However, the reaction quotient should actually be the reciprocal of that. So Q = 0.1 M / x.

Now, let's put on our funny hats and solve for x:

0.024 V = 0 V - (8.314 J/(mol·K) * 298 K) / (2 * 96485 C/mol) * ln(0.1 M / x)

Now, let's simplify that and solve:

ln(0.1 M / x) = -0.024 V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K)
ln(0.1 M / x) = -944.84
0.1 M / x = e^(-944.84)
x = 0.1 M / e^(944.84)

Now, here's the punchline: the value of e^(944.84) is quite large, so we're talking about a minuscule concentration of Zn2+. In other words, it's probably better to switch to scientific notation to express it! So, don't be shocked when I tell you that your calculated concentration is really close to zero.

But let's not be disappointed, chemistry can be quite unpredictable! Double-check your calculations just to be sure. And remember, with a little more voltage and humor, anything is possible!

To calculate the concentration of Zn2+(aq) at the cathode, we can use the Nernst equation:

E = E° - (0.0592/n)*log(Q)

Where:
E is the cell potential
E° is the standard cell potential
n is the number of electrons transferred in the balanced equation
Q is the reaction quotient

In this case, the balanced equation for the concentration cell is:

Zn(s) + Zn2+(aq, 0.100 M) -> Zn2+(aq, x M) + Zn(s)

The standard cell potential for this concentration cell is zero since the concentrations on both sides are the same. Therefore, we can simplify the equation to:

E = - (0.0592/n)*log(Q)

Given that the cell potential is 24.0 mV (or 0.024 V) and temperature is 25°C (or 298 K), we'll use these values in our calculation.

E = 0.024 V
T = 298 K

Now, we need to determine the reaction quotient, Q. The reaction quotient is the ratio of the concentrations of the species involved:

Q = [Zn2+(aq, x M)] / [Zn2+(aq, 0.100 M)]

Plugging in the values into the Nernst equation, we have:

0.024 = - (0.0592/n)*log([Zn2+(aq, x M)] / [Zn2+(aq, 0.100 M)])

Simplifying further:

- log([Zn2+(aq, x M)] / [0.100]) = - (0.024 * n) / (0.0592)

Taking the antilog of both sides:

[Zn2+(aq, x M)] / [0.100] = 10^(- (0.024 * n) / (0.0592))

[Zn2+(aq, x M)] / [0.100] = 10^(-0.4081 * n)

Multiplying both sides by 0.100:

[Zn2+(aq, x M)] = 0.100 * 10^(-0.4081 * n)

Since we know that n is 2 electrons (from the balanced equation), we can substitute it into the equation:

[Zn2+(aq, x M)] = 0.100 * 10^(-0.4081 * 2)

[Zn2+(aq, x M)] = 0.100 * 10^(-0.8162)

[Zn2+(aq, x M)] = 0.100 * 0.1712

[Zn2+(aq, x M)] = 0.01712 M

So, the concentration of Zn2+(aq) at the cathode is approximately 0.01712 M.

To find the concentration of the Zn2+ ion at the cathode in the zinc concentration cell, you can use the Nernst equation:

E = E° - (RT/nF) * ln(Q)

Where:
- E is the measured cell potential (24.0 mV or 0.024 V)
- E° is the standard cell potential (0 V in this case since both electrodes are the same)
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25°C = 298 K)
- n is the number of moles of electrons transferred in the cell (2 in this case since Zn2+(aq) gains 2 electrons during reduction)
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient, which is the ratio of concentrations of the Zn2+ ions at the anode and cathode.

Since the anode and cathode are both zinc electrodes, the concentration of Zn2+ ions at both sides is the same, so Q = [Zn2+]/[Zn2+].
Let's call the unknown concentration of Zn2+ at the cathode as x M.

Now, let's substitute the known values into the equation and solve for x:

0.024 = 0 - (8.314 * 298) / (2 * 96485) * ln(x/0.1)

Simplifying this equation will give you:

0.024 = -0.012839 * ln(x/0.1)

Rearranging the equation to isolate x:

ln(x/0.1) = -0.024 / -0.012839
x/0.1 = e^(-0.024 / -0.012839)

Taking the inverse natural logarithm and multiplying both sides by 0.1 yields:

x = 0.1 * e^(-0.024 / -0.012839)

Evaluating this expression will give you the concentration of Zn2+ ions at the cathode.

x ≈ 0.101 M

Therefore, the estimated concentration of Zn2+ ions at the cathode is approximately 0.101 M.

That looks ok to me although I seem to remember 0.7 something when I scratched this out a day or so ago but I've thrown my old worksheets away. I usually use

E = (-0.05916/2)log(0.1/x) but since you have changed the - sign in front to + then you have changed to x/0.1 and that makes it ok again.