Calculate the standard cell potential given the following standard reduction potentials:
Al3++3e−→Al;E∘=−1.66 V
Fe2++2e−→Fe;E∘=−0.440 V
To calculate the standard cell potential, we use the Nernst equation:
Ecell = E°cell - (0.0592/n)log(Q)
where:
Ecell is the standard cell potential
E°cell is the standard reduction potential of the cell
n is the number of electrons transferred in the cell reaction
Q is the reaction quotient, which is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficient
In this case, we can use the given reduction half-reactions directly as the reactions occurring in the cell:
Al3+ + 3e- → Al, with E° = -1.66 V
Fe2+ + 2e- → Fe, with E° = -0.440 V
Since there are no coefficients listed for these half-reactions, we can assume they have balanced stoichiometry coefficients of 1.
The overall cell reaction is the sum of these half-reactions:
2Al3+ + 6e- → 2Al
3Fe2+ + 6e- → 3Fe
By multiplying these half-reactions as needed to balance the number of electrons transferred, we get:
2Al3+ + 3Fe2+ → 2Al + 3Fe
Now, we can calculate the standard cell potential:
Ecell = E°cell - (0.0592/n)log(Q)
Since the overall reaction involves the transfer of 6 electrons, n = 6.
The reaction quotient Q is the ratio of the concentrations of products (Al and Fe) to the concentrations of reactants (Al3+ and Fe2+). However, since Q is not specified in the question, we cannot calculate the exact value of Ecell without additional information about the concentrations.
The standard cell potential can be calculated once Q is known.
To calculate the standard cell potential, you need to apply the equation:
E°cell = E°cathode - E°anode.
In this case, the Al3+ / Al reduction half-reaction will be the cathode (reduction) half-reaction, and the Fe2+ / Fe reduction half-reaction will be the anode (oxidation) half-reaction.
Given that E°cathode = -1.66 V and E°anode = -0.440 V, you can substitute the values into the equation to find the standard cell potential:
E°cell = E°cathode - E°anode
E°cell = (-1.66 V) - (-0.440 V)
E°cell = -1.22 V
Therefore, the standard cell potential for the given reduction potentials is -1.22 V.
You don't give enough information to know if you want the Eocell for a spontaneous or non-spontaneous cell. I assume you want the spontaneous reaction.
Al(s) ==> Al^3+ + 3e Eo = +1.66
Fe^2+ + 2e ==> Fe(s) Eo = -0.440
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2Al + 3Fe^2+ ==> 2Al^3+ + 3Fe
Add Eo values to find Eocell.