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
There are a couple of ways of doing this. I think the easier of the two is as follows:
ClO2 + e ==> ClO2^- Eo = +0.95v
Sn(s) ==> Sn^2+ + 2e Eo = -0.14
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2ClO2 + Sn(s) ==>2ClO2^- + Sn^2+ which is the cell rxn at standard states 1 atm, 1M, etc. Eocell = 0.95+(-0.14) = ?
Then Ecell = Eocell -(0.0592/n)*log Q
where Q = (ClO2^-)^2*(Sn^2+)/p^2ofClO2*(Sn)(s)(1.74)(1.44)^2/(0.120)^2 * 1
Well, if we have an electrochemical cell, we can calculate the cell potential by subtracting the reduction potential of the anode from the reduction potential of the cathode. So in this case, we have:
Ecell = Ered - Eox
Since Ered is 0.95 and Eox is -0.14, we can plug those values into the equation:
Ecell = 0.95 - (-0.14)
Ecell = 1.09
So at a temperature of 25°C, the cell potential of this electrochemical cell is 1.09 volts. Just be careful not to mix up your volts with clowns!
To calculate the cell potential at 25°C, we can use the Nernst equation, which is given as:
Ecell = Ecathode - Eanode
Where Ecathode is the reduction potential of the cathode and Eanode is the reduction potential of the anode.
Given:
E Sn>Sn2+ = -0.14 V (cathode reduction potential)
E ClO2 = 0.95 V (anode reduction potential)
Now, we need to balance the half-reactions and determine the overall reaction.
Balanced half-reactions:
Ox: Sn(s) → Sn2+(aq) + 2 e-
Red: ClO2(g) + e- → ClO2-(aq)
To balance the half-reactions, we need to multiply the first half-reaction by 2 to ensure that the electrons cancel out.
Balanced half-reactions:
Ox: 2 Sn(s) → 2 Sn2+(aq) + 4 e-
Red: ClO2(g) + e- → ClO2-(aq)
Now, we can write the overall reaction by summing up the two half-reactions:
Overall reaction: 2 Sn(s) + ClO2(g) → 2 Sn2+(aq) + ClO2-(aq)
Now, let's calculate the cell potential using the Nernst equation:
Ecell = Ecathode - Eanode
Ecell = 0.95 V - (-0.14 V)
Ecell = 1.09 V
Therefore, the cell potential at 25°C is 1.09 volts.
To calculate the cell potential at 25°C, we need to use the Nernst equation, which relates the cell potential to the standard cell potential and the concentrations of the species involved.
The Nernst equation is given as:
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 electrons transferred in the balanced equation
- F is Faraday's constant (96,485 C/mol)
- Q is the reaction quotient
First, let's determine the balanced overall reaction by summing the half-reactions:
Sn(s) + ClO2(g) → Sn2+(aq) + ClO2-(aq)
The balanced equation shows that n = 1, as one electron is transferred.
Next, let's calculate the reaction quotient (Q) using the concentrations of the species involved.
Q = [Sn2+][ClO2-] / [Sn][ClO2]
= (1.74 M * 1.44 M) / (1.0 M * 0.120 atm)
Now we have all the values required to use the Nernst equation and calculate the cell potential.
Ecell = E°cell - (RT/nF) * ln(Q)
Plugging in the known values:
- E°cell = E°red - E°ox (since E°cell = E°red - E°ox)
- E°red = 0.95 V (given)
- E°ox = -0.14 V (given)
- R = 8.314 J/(mol·K)
- T = 25°C = 298 K
- n = 1
- F = 96,485 C/mol
- Q = calculated earlier
Ecell = (0.95 V) - ((8.314 J/(mol·K)) * (298 K) / (1 * 96,485 C/mol)) * ln(Q)
By substituting the values and solving the equation, you can calculate the cell potential at 25°C.