The voltage generated by the zinc concentration cell described by,
Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)
is 23.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.
Ecell = -(0.0592/n)log(dilute soln/concd soln). Substitute to obtain
0.023 = -(0.0592/2)log(x/0.1) and solve for x. You know to let x = dilute soln because the voltage is +.
Everything on the above equation is correct besides the log part. its supposed to be log(.1/x), not log(x/.1)...
To calculate the concentration of Zn2+ ion at the cathode, we can use the Nernst equation. The Nernst equation relates the cell potential to the concentration of the ions involved in the cell reaction.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E = cell potential
E° = standard cell potential
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
n = number of moles of electrons transferred in the balanced cell reaction
F = Faraday's constant (96485 C/mol)
Q = reaction quotient
In the given zinc concentration cell, the reaction at the cathode is:
Zn2+ (aq) + 2e- -> Zn (s)
Since zinc is a solid, its concentration does not affect the reaction quotient, and therefore, the concentration of Zn2+ ion at the cathode is unknown, denoted by "? M". This is the concentration we need to calculate.
The standard cell potential (E°) for this cell can be obtained from standard reduction potentials. The standard reduction potential for the zinc half-reaction can be found in a table. For this cell, E° = -0.763 V.
Now, let's substitute the given values and solve for the unknown concentration:
E = 23.0 mV = 0.023 V (since 1 V = 1000 mV)
E° = -0.763 V
R = 8.314 J/(mol*K)
T = 25 °C = 298 K
n = 2 (since 2 moles of electrons are transferred in the balanced cell reaction)
F = 96485 C/mol
Plugging these values into the Nernst equation, we have:
0.023 = -0.763 - (8.314 * 298 / (2 * 96485)) * ln(Q)
Simplifying and rearranging the equation, we get:
ln(Q) = -(0.023 + 0.763) * (2 * 96485) / (8.314 * 298)
ln(Q) = -0.033
Now, we can take the natural logarithm (ln) of Q:
Q = e^(-0.033)
Q ≈ 0.967
Finally, we can calculate the concentration of Zn2+ ion at the cathode using the reaction quotient:
Q = [Zn2+ (aq)] / [Zn2+ (aq, 0.100 M)]
0.967 = [Zn2+ (aq)] / 0.100
[Zn2+ (aq)] = 0.967 * 0.100
[Zn2+ (aq)] ≈ 0.097 M
Therefore, the concentration of Zn2+ ion at the cathode is approximately 0.097 M.