What is the cell potential for the reaction
Mg(s) + Fe^2+(aq) -----> Mg^2+(aq) + Fe (s)
at 63 degrees celsius when [Fe^2+]= 3.30 M and [Mg^2+]= 0.110 M
Find the Eo for the cell as written. Then Ecell =
Eocell + (RT/nF)lnQ
Q = (Mg^2+)/(Fe(2+)
Substitute and solve for Ecell.
To find the cell potential for the given reaction, you can use the Nernst equation. The Nernst equation relates the cell potential to the concentrations of the species involved in the reaction.
The Nernst equation is given by:
Ecell = E°cell - (0.0592 V/n) * log(Q)
where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- Q is the reaction quotient
- n is the number of electrons transferred
First, let's find the standard cell potential (E°cell). The standard cell potential is a measure of the tendency of the reaction to occur under standard conditions. You can find tabulated values of standard cell potentials for different reactions.
In this case, the given reaction involves the reduction of Fe^2+ to Fe(s). The standard cell potential for this half-reaction can be found in a standard reduction potential table. Let's assume the standard cell potential for this half-reaction is +0.44 V.
Next, we need to calculate the reaction quotient (Q). The reaction quotient is the ratio of the concentrations of the products over the reactants, each raised to the power of their stoichiometric coefficients.
In this case, the reaction quotient (Q) can be calculated as:
Q = [Mg^2+]/[Fe^2+]
Substituting the given values, Q = 0.110 M / 3.30 M = 0.0333
Now, we can substitute the values obtained into the Nernst equation to calculate the cell potential (Ecell):
Ecell = +0.44 V - (0.0592 V/n) * log(0.0333)
Since the given reaction involves the transfer of two electrons, n = 2. We can substitute this value into the equation:
Ecell = +0.44 V - (0.0592 V/2) * log(0.0333)
Now, you can solve this equation to find the cell potential (Ecell) at 63 degrees Celsius.