Chapter 17: Problem 73
Consider the cell described below: $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M)\right|\left|\mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{mol} / \mathrm{L}\). (Assume \(T=25^{\circ} \mathrm{C} .\) )
Short Answer
Step by step solution
Write down the balanced redox equation.
Find the standard cell potential.
Determine the changes in the concentrations of ions in the reaction.
Apply the Nernst equation to calculate the cell potential after changes in the concentrations of the ions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst Equation
\[\[\begin{align*}E_{cell} = E^{0}_{cell} - \frac{RT}{nF} \times ln (Q) \end{align*}\]\]
where
- \(E_{cell}\) is the cell potential under non-standard conditions,
- \(E^{0}_{cell}\) represents the standard cell potential,
- \(R\) is the universal gas constant (8.314 J/mol·K),
- \(T\) is the temperature in Kelvin,
- \(n\) is the number of moles of electrons exchanged in the redox reaction,
- \(F\) is Faraday's constant (96485 C/mol), and
- \(Q\) is the reaction quotient, a dimensionless number representing the ratio of concentrations of products to reactants at any given point in the reaction.
Redox Reaction
\[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}\]
Here, zinc metal (Zn) is oxidized to zinc ions (\(Zn^{2+}\)), losing two electrons in the process, while copper ions (\(Cu^{2+}\)) are reduced to copper metal (Cu), each gaining two electrons. This electron exchange is the fundamental process driving the cell's electrical output.
Standard Reduction Potentials
Reaction Quotient
\[Q = \frac{[\text{Products}]}{[\text{Reactants}]}\]
Since our reaction has zinc ions as the product and copper ions as the reactant, the reaction quotient can be written as:\(Q = \frac{[Zn^{2+}]}{[Cu^{2+}]}\). As the reaction proceeds past the initial state, the concentrations of zinc and copper ions change, affecting the cell potential. The Nernst equation accommodates these concentration changes and thus enables the calculation of the cell potential at any given state of the reaction.