Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.859 M and [Ni2 ] = 0.0190 M. Standard reduction potentials can be found here.
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To calculate the cell potential for the reaction at 25.00 °C, we need to use the Nernst equation. The Nernst equation relates the standard cell potential to the concentrations of the species involved in the reaction.
First, let's write the balanced half-reactions for the reaction:
Cr2+(aq) + 2e- -> Cr(s)
Ni2+(aq) + 2e- -> Ni(s)
The standard reduction potential for the Cr2+(aq) + 2e- -> Cr(s) half-reaction is given as -0.91 V.
The standard reduction potential for the Ni2+(aq) + 2e- -> Ni(s) half-reaction is given as -0.25 V.
Now, we can use the Nernst equation:
E = E° - (RT / nF) * ln(Q)
Where:
E is the cell potential
E° is the standard cell potential
R is the ideal gas constant (8.314 J/K·mol)
T is the temperature in Kelvin
n is the number of moles of electrons transferred (in this case, 2)
F is the Faraday constant (96485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
First, let's calculate the reaction quotient (Q) using the given concentrations:
Q = [Cr2+] / [Ni2+]
Q = 0.859 M / 0.0190 M
Q = 45.21
Now, let's calculate the cell potential (E) using the Nernst equation:
E = -0.91 V - ((8.314 J/K·mol) * (25.00 + 273.15 K) / (2 * 96485 C/mol)) * ln(45.21)
Using the natural logarithm (ln) function available on most calculators, we can solve for E:
E ≈ -0.91 V - (0.0257 V) * ln(45.21)
E ≈ -0.91 V - (0.0257 V) * 3.813
E ≈ -0.91 V - 0.0981 V
E ≈ -1.01 V
Therefore, the cell potential for the given reaction at 25.00 °C is approximately -1.01 V.