Consider the following electochemical cell.
Pt | Fe3+ (0.0600 M), Fe2+ (6.50e-5 M) || Sn2+ (0.0620 M), Sn4+ (3.90e-4 M) | Pt
(a) Calculate the voltage of the cell, Ecell, including the sign.
Use the Nernst equation to calculate the half cell potential of the Fe+3/Fe+2 electrode.
Use the Nernst equation to calculate the half cell potential of the Sn+4/Sn+2 electrode. Note I used REDUCTIONS for both. Then I would proceed as follows:
Identify the more negative electrode of the two reduction potentials, reverse it and the sign, add the two half equations to find the SPONTANEOUS cell reaction and spontaneous cell potential. Compare the spontaneous reaction with the reaction of the cell, which I've given below. Adjust your equation and potential (the sign) as needed. The reaction of the cell, as written, is:
2Fe^+3 + Sn^+2 ==> 2Fe+2 + Sn^+4
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To calculate the voltage of the cell (Ecell), you need to use the Nernst equation. The Nernst equation relates the voltage of a cell to the concentrations of the species involved in the cell reaction. The Nernst equation is written as follows:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- E°cell: Standard cell potential (constant for a given reaction)
- R: Gas constant (8.314 J/(mol·K))
- T: Temperature in Kelvin (K)
- n: Number of electrons transferred in the balanced cell reaction
- F: Faraday constant (96485 C/mol)
- Q: Reaction quotient (ratio of product concentrations to reactant concentrations)
In this case, you already have all the necessary information except for E°cell. E°cell can be calculated by subtracting the reduction half-reaction potential from the oxidation half-reaction potential.
Now, let's calculate the voltage of the given cell.
First, write the two half-reactions occurring in this cell:
Oxidation half-reaction: Fe3+ + e- -> Fe2+
Reduction half-reaction: Sn4+ + 2e- -> Sn2+
The standard reduction potentials (E° values) for these half-reactions can be looked up in a standard reduction potential table.
E°cell = E°reduction - E°oxidation
Now, substitute the E° values into the equation and solve for E°cell:
E°cell = E°reduction - E°oxidation
E°cell = (E°reduction of Sn2+/Sn4+) - (E°oxidation of Fe3+/Fe2+)
After calculating the values of E°cell, substitute them into the Nernst equation along with the known concentrations of Fe3+, Fe2+, Sn2+, and Sn4+ to get the Ecell of the cell. Remember to include the appropriate signs for the concentrations of Fe3+, Fe2+, Sn2+, and Sn4+ in the reaction quotient (Q).
Finally, once you've calculated the value of Ecell, include the sign (positive or negative) to indicate the direction of electron flow in the cell.