Determine the cell potential for Ni(s) + Fe²⁺(aq) → Ni²⁺(aq) + Fe(s) where [Ni²⁺] = 0.60 M and [Fe²⁺] = 0.0030 M using the following standard reduction potentials. Ni²⁺(aq) + 2e⁻ → Ni(s) E° = -0.25 V and Fe²⁺(aq) + 2e⁻ → Fe(s) E° = -0.44 V.
To determine the cell potential for the given reaction, you can use the Nernst equation. The Nernst equation relates the cell potential (E) to the standard reduction potentials (E°) and the concentrations of the reactants. The equation is as follows:
E = E° - (RT / nF) * ln(Q)
where:
E is the cell potential
E° is the standard reduction potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of electrons transferred in the balanced equation
F is the Faraday constant (96,485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
First, let's write the balanced equation for the reaction:
Ni(s) + Fe²⁺(aq) → Ni²⁺(aq) + Fe(s)
From the balanced equation, we can see that n = 2 because two electrons are transferred in the reaction.
Next, calculate the reaction quotient (Q) using the given concentrations:
Q = ([Ni²⁺] / [Fe²⁺])²
Substituting the values, we get:
Q = (0.60 M / 0.0030 M)² = 120²
Now, substitute the given standard reduction potentials and the calculated reaction quotient into the Nernst equation:
E = (-0.25 V) - ((8.314 J/(mol·K) * T) / (2 * (96,485 C/mol)) * ln(120²)
The temperature (T) should be specified to complete the calculation. Plug in the temperature value in Kelvin and solve the equation to determine the cell potential (E).