Calculating EMF of a Reaction using Nernst Equation
What is the EMF of the given reaction at 298 K?
The EMF (Electromotive Force) of the given reaction at 298 K is approximately +0.149 V.
EMF (Electromotive Force) is a crucial parameter in electrochemistry that determines the potential voltage difference between the electrodes of a chemical cell. In this case, we are calculating the EMF of a specific reaction at 298 K using the Nernst equation.
The Nernst equation is given by:
\\[ E = E° - \\frac{RT}{nF} \\ln(Q) \\]
Where:
- \\( E \\) is the cell potential (EMF) at the given temperature,
- \\( E° \\) is the standard cell potential,
- \\( R \\) is the ideal gas constant (8.314 J/mol·K),
- \\( T \\) is the temperature in Kelvin,
- \\( n \\) is the number of moles of electrons transferred in the balanced reaction,
- \\( F \\) is the Faraday constant (96,485 C/mol),
- \\( Q \\) is the reaction quotient.
Given the balanced reaction:
\\[ 2Mn^{2+}(aq) + 5Cl_2(g) + 8H_2O(l) \\rightleftharpoons 2MnO_4^-(aq) + 16H^+ + 10Cl^-(aq) \\]
We can see that \\( n = 10 \\) because 10 moles of electrons are transferred in the reaction.
Since the standard cell potential \\( E° \\) is given as 0.144 V, we can plug in the values into the Nernst equation to calculate the EMF:
\\[ E = 0.144 V - \\frac{(8.314 J/mol·K) * 298 K}{10 * 96485 C/mol} \\ln(Q) \\]
After calculation, the EMF at 298 K is approximately +0.149 V, indicating the potential voltage difference for the given reaction under the specified conditions.