Calculate the cell potential for the following reaction as written at 51 degrees C, given that [Zn2+] = 0.837 M and [Sn2+] = 0.0180 M.
Zn(s) + Sn^2+(aq) = Zn^2+(aq)+Sn(s)
E = __ V
Reduction Half-Reaction Standard Potential Ered° (V)
Zn2+(aq) + 2e– → Zn(s) –0.76
Sn2+(aq) + 2e– → Sn(s) –0.14
I was hoping someone could tell me if this is correct:
-0.76 - 0.14 = -0.90 V
(-0.9) - (8.314 * 324 K)/(2 * 96485) * ln[0.837/0.0180]
= (-0.9) - 0.053596262
= -0.95359 V?
To calculate the cell potential (E) for the given reaction at 51 degrees Celsius, you need to use the Nernst equation. The Nernst equation relates the standard cell potential (Ered°) to the cell potential under non-standard conditions (E):
E = Ered° - (RT/nF) * ln(Q)
Where:
- E is the cell potential under non-standard conditions.
- Ered° is the standard cell potential.
- R is the ideal gas constant (8.314 J/(mol·K)).
- T is the temperature in Kelvin (51°C = 324 K).
- n is the number of moles of electrons transferred in the balanced equation (2 in this case).
- F is the Faraday constant (96485 C/mol).
- Q is the reaction quotient, which is calculated as the ratio of product concentrations over reactant concentrations, each raised to the power of their stoichiometric coefficients.
Given:
- Ered° for Zn2+/Zn = -0.76 V
- Ered° for Sn2+/Sn = -0.14 V
- [Zn2+] = 0.837 M
- [Sn2+] = 0.0180 M
The balanced equation is: Zn(s) + Sn^2+(aq) = Zn^2+(aq) + Sn(s)
First, calculate the reaction quotient Q:
Q = ([Zn2+]/[Sn2+])^2
Q = (0.837/0.0180)^2 = 253.690
Now, substitute the values into the Nernst equation:
E = Ered° - (RT/nF) * ln(Q)
E = (-0.76 V) - ((8.314 J/(mol·K)) * (324 K)) / (2 * 96485 C/mol) * ln(253.690)
E = -0.76 V - 0.053596262 V
E = -0.813596262 V
Therefore, the cell potential (E) for the given reaction at 51 degrees Celsius is approximately -0.8136 V.
To calculate the cell potential for the given reaction, you need to use the Nernst equation, which relates the cell potential to the standard potential and the concentrations of the reactants involved.
The Nernst equation is given as:
E = E° - (RT/nF) * ln(Q)
Where:
E is the cell potential,
E° is the standard potential,
R is the ideal gas constant (8.314 J/mol·K),
T is the temperature in Kelvin,
n is the number of moles of electrons transferred,
F is the Faraday constant (96,485 C/mol), and
Q is the reaction quotient.
First, let's calculate the reaction quotient Q:
Q = [Zn^2+]/[Sn^2+]
= 0.837/0.018
= 46.5
Now we can substitute the values into the Nernst equation:
E = (-0.9) - (8.314 * (51 + 273.15)) / (2 * 96485) * ln(46.5)
Let's calculate the right-hand side of the equation step by step:
ln(46.5) ≈ 3.841
(8.314 * (51 + 273.15)) / (2 * 96485) ≈ 0.0506
Substituting these values back into the equation:
E ≈ (-0.9) - (0.0506 * 3.841)
≈ -0.9 - 0.1946
≈ -1.0946 V
Therefore, the cell potential for the given reaction at 51 degrees C is approximately -1.0946 V.
It isn't correct although I didn't finish working the problem.
Zn ==> Zn^2+ + 2e = +0.76
Sn^2+ + 2e ==> Sn = -=0.14
Therefore, for the rxn
Zn + Sn^2+ ==> Sn + Zn^2+ E = 0.76+(-0.14) = 0.62 v.
I changed the Zn to an oxidation, left the Sn as a reduction, added the oxidn half to the redn half and added the two corresponding E values to find Ecell.
I think the rest is ok if you change the 0.9 to 0.62. By the way reporting to 5 significant figures is ridiculous.