An electrochemical cell is based on these two half-reactions:
Ox: Sn(s) → Sn2+(aq, 1.74 M) + 2 e-
Red: ClO2(g, 0.120 atm) + e- → ClO2-(aq, 1.44 M)
Calculate the cell potential at 25°C
E Sn>Sn2+ = -.14
E ClO2 = .95
To calculate the cell potential at 25°C, we need to use the Nernst equation, which relates the cell potential to the concentrations and standard reduction potentials of the species involved.
The Nernst equation is given by:
Ecell = E°cell - (RT / nF) * ln(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of moles of electrons transferred in the balanced equation
F is Faraday's constant (96485 C/mol)
ln(Q) is the natural logarithm of the reaction quotient
In this case, the cell potential can be calculated by considering the half-reactions:
Ox: Sn(s) → Sn2+(aq, 1.74 M) + 2 e-
Red: ClO2(g, 0.120 atm) + e- → ClO2-(aq, 1.44 M)
From the given information, we have the standard reduction potentials:
E° Sn > Sn2+ = -0.14 V
E° ClO2 = 0.95 V
First, let's calculate the reaction quotient Q. This can be done using the concentrations and partial pressure of the species involved:
Q = [Sn2+] / ([ClO2-] * [ClO2])
Given concentrations and partial pressure:
[Sn2+] = 1.74 M
[ClO2-] = 1.44 M
[ClO2] = 0.120 atm
Next, we substitute the values into the Nernst equation:
Ecell = E°cell - (RT / nF) * ln(Q)
Assuming the number of electrons transferred (n) is 2, and the gas constant (R) is 8.314 J/(mol·K), the temperature (T) is 25°C which is 298 K, and Faraday's constant (F) is 96485 C/mol, we can calculate the cell potential (Ecell) by plugging in the values.
Ecell = (0.95 V) - ((8.314 J/(mol·K)) * (298 K) / (2 * 96485 C/mol)) * ln(Q)
Simplifying the equation and substituting the value of ln(Q), we can calculate the cell potential.
Ecell = 0.95 V - (0.003427 V) * ln(Q)
Once you have calculated the value of the natural logarithm of Q (ln(Q)), you can substitute it into the equation to find the cell potential (Ecell).