Chapter 6: Problem 43
For a cell reaction involving a two electron, the standard EMF of the cell is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction at \(25^{\circ} \mathrm{C}\) will be a. \(1 \times 10^{-10}\) b. \(29.5 \times 10^{-2}\) c. 10 d. \(1 \times 10^{10}\)
Short Answer
Step by step solution
Understand the Nernst Equation
Convert Temperature and Constants
Substitute Values into Nernst Equation
Calculate \( \ln K \) and Solve for \( K \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst equation
The standard form of the Nernst equation is given by:
- \(E = E^0 - \frac{RT}{nF} \ln Q\)
- \(E\) is the cell potential at non-standard conditions.
- \(E^0\) is the standard EMF of the cell.
- \(R\) is the universal gas constant \(8.314 \, \text{J/mol} \cdot \text{K}\).
- \(T\) is the temperature in Kelvin.
- \(n\) is the number of moles of electrons transferred in the reaction.
- \(F\) is the Faraday constant \(96485 \, \text{C/mol}\).
- \(Q\) is the reaction quotient, similar to the equilibrium constant but not at equilibrium.
The Nernst equation is crucial for understanding how cell potentials change with varying reactant concentrations or with changing conditions.
standard EMF
To calculate the standard EMF of a cell, we use the standard reduction potentials derived from the half-reactions of the redox pairs. The equation is:
- \(E^0 = E^0_{\text{cathode}} - E^0_{\text{anode}}\)
- \(E^0_{\text{cathode}}\) is the standard reduction potential for the reduction half-reaction at the cathode.
- \(E^0_{\text{anode}}\) is the standard reduction potential for the oxidation half-reaction at the anode.
A positive standard EMF indicates that a redox reaction is spontaneous in the forward direction.
redox reaction
In an oxidation half-reaction, a substance loses electrons, increasing its oxidation state. Conversely, in a reduction half-reaction, a substance gains electrons, reducing its oxidation state.
For example, consider the reaction between zinc and copper(II) ions:
- Oxidation: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-\)
- Reduction: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)
- \(\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}\)
Moreover, balancing redox reactions allows us to comprehend the stoichiometry behind these processes properly and understand the material flow during a reaction.
The interplay between oxidation and reduction in redox reactions is at the heart of electrochemical cells and numerous natural and industrial processes.