I need help determining a concentration. A voltaic cell has a Hg/Hg2^2+ electrode and a Sn/Sn^2+ electrode (Enaught cell = -0.14V), Ecell = 1.04V at 298K. If [Hg2^2+]=0.24M calculate[Sn2+]. I used this formula, Ecell=Enaught Cell - (0.0592/n)logQ. my Q ends up as log(0.24M/[Sn2+]). How do I solve for the concentration of Sn2+?
To solve for the concentration of Sn^2+ in the voltaic cell, you will need to rearrange the equation:
Ecell = E°cell - (0.0592/n) * logQ
Given that E°cell = -0.14V and Ecell = 1.04V, we can substitute these values into the equation:
1.04V = -0.14V - (0.0592/n) * logQ
Next, isolate the logQ term:
(0.0592/n) * logQ = 1.04V - (-0.14V)
(0.0592/n) * logQ = 1.18V
Then, divide both sides of the equation by (0.0592/n):
logQ = (1.18V) * (n/0.0592)
At this point, we need to calculate the value of n, which represents the number of electrons exchanged during the reaction. The balanced equation for the reaction at the Sn/Sn^2+ electrode is:
Sn + 2e^- → Sn^2+
Since two electrons are involved in the reaction, n = 2.
Now we can substitute the value of n into the equation:
logQ = (1.18V) * (2/0.0592)
Simplifying further:
logQ = 39.73
To solve for Q, we need to take the antilog of both sides:
Q = 10^(39.73)
Now we can substitute the values into the expression for Q:
Q = 10^(log(0.24M/[Sn^2+]))
Since log and 10^ are inverse operations, we can rewrite the expression as:
Q = 0.24M / [Sn^2+]
Finally, isolate [Sn^2+] by rearranging the equation:
[Sn^2+] = 0.24M / Q
Substitute the value of Q calculated earlier:
[Sn^2+] = 0.24M / 10^(39.73)
Calculating the value:
[Sn^2+] ≈ 2.40 × 10^(-40) M
Therefore, the concentration of Sn^2+ in the voltaic cell is approximately 2.40 × 10^(-40) M.