Calculate the cell potential for the following reaction as written at 53 °C, given that [Zn2 ] = 0.862 M and [Ni2 ] = 0.0110 M. Standard reduction and oxidation potentials can be found here.
Zn(s) + Ni^2+(aq) --> Zn^2+(aq) + Ni(s)
To calculate the cell potential for the given reaction at 53°C, we need to use the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential and the concentrations of the species involved. The equation is as follows:
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 redox equation
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient
Step-by-step solution:
1. First, let's identify the half-reactions involved in the overall reaction:
Oxidation half-reaction: Zn(s) → Zn^2+(aq) + 2e^-
Reduction half-reaction: Ni^2+(aq) + 2e^- → Ni(s)
2. Look up the standard reduction potentials for the half-reactions at 298K from the provided source. The standard reduction potentials are as follows:
E°(Zn^2+/Zn) = -0.76 V
E°(Ni^2+/Ni) = -0.25 V
3. The overall standard cell potential (E°cell) is determined by subtracting the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction:
E°cell = E°(Ni^2+/Ni) - E°(Zn^2+/Zn)
E°cell = -0.25 V - (-0.76 V)
E°cell = 0.51 V
4. Convert the temperature to Kelvin:
53°C + 273.15 = 326.15 K
5. Calculate the reaction quotient (Q) using the concentrations of the species involved from the given information:
Q = [Zn^2+]/[Ni^2+]
Q = 0.862 M / 0.0110 M
Q = 78.36
6. Now, substitute the values into the Nernst equation and solve for Ecell:
Ecell = E°cell - (RT/nF) * ln(Q)
= 0.51 V - ((8.314 J/(mol*K)) * (326.15 K) / (2 * 96485 C/mol)) * ln(78.36)
Calculating this expression will give you the cell potential (Ecell) for the given reaction at 53°C.