Calculate the cell potential for the following Galvanic cell:
Co(s) Co2+ (0.15 mol L-1, ag)Fe 2+ (0.68 mol L-1, ag)Fe(s)
Given that the standard reduction potential Eo for Co2+ is -0.28 V and Eo for Fe2+ is -0.44 V.
Anwer is -0.14 V
The left side of the above equaton is the annode and the right the cathode therefore -0.44 - -0.28 = -0.16 which is incorrect. What steps are taken showing maths and answer of -0.14. Are charges included?
maths to get -0.14 V.
To calculate the cell potential for the given galvanic cell, you can use the Nernst equation. The Nernst equation relates the cell potential to the concentrations of the species involved in the half-reactions.
The balanced equation for the cell is as follows:
Co(s) + Fe2+(aq) → Co2+(aq) + Fe(s)
First, let's calculate the standard cell potential (E°cell) using the standard reduction potentials for Co2+ and Fe2+.
E°cell = E°cathode - E°anode
E°cell = E°Fe/Fe2+ - E°Co/Co2+
= -0.44 V - (-0.28 V)
= -0.16 V
Now, to calculate the cell potential under non-standard conditions, we use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
Ecell = cell potential under non-standard conditions
E°cell = standard cell potential
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of moles of electrons transferred in the overall balanced equation
F = Faraday's constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient
In this case, the reaction quotient (Q) can be calculated as follows:
Q = ([Co2+]/[Fe2+])
Now, let's substitute the given concentrations into the Nernst equation:
Ecell = -0.16 V - [(8.314 J/(mol·K)) * (298 K) / (2 mol e^- * 96485 C/mol)] * ln([0.15 mol/L] / [0.68 mol/L])
Ecell = -0.16 V - (0.0257 V) * ln(0.22)
Ecell = -0.14 V
So, the cell potential for the given galvanic cell is -0.14 V under the given concentrations and conditions.
Charges are not directly included in this calculation because reduction potentials are given per mole of electrons transferred. However, the stoichiometric coefficients in the balanced equation reflect the number of electrons transferred in the half-reactions. In this case, it is 2 mol e^-.