Question

For $Zn(s)+Cu^{2+}(0.1M) \rightarrow Zn^{2+}(0.001 M)+Cu(s)$, calculate $E_{cell}$ if $E^0_{cell} = 1.1V$. ________.

• A. 1.218 V
• B. 1.118 V
• C. 1.159 V
• D. 1.041 V
• E. 0.982 V

Hint:

If product concentration $<$ reactant concentration, $E_{cell} > E^0_{cell}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This problem requires the application of the Nernst equation, which describes how the electric potential of an electrochemical cell (\(E_{cell}\)) deviates from its standard potential (\(E^{\circ}_{cell}\)) as a function of the concentrations of the reactants and products.
Step 2: Key Formula or Approach
The Nernst equation is given by: \[ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF}\ln Q \] At standard temperature (298 K) and converting to base-10 logarithm, the equation becomes: \[ E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n}\log_{10} Q \] where: - \(n\) is the number of moles of electrons transferred in the balanced reaction. - \(Q\) is the reaction quotient.
Step 3: Detailed Explanation
1. Determine the number of electrons transferred (n).
The overall reaction is \(Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)\). The process can be broken into half-reactions: - Oxidation: \(Zn \to Zn^{2+} + 2e^-\) - Reduction: \(Cu^{2+} + 2e^- \to Cu\) From the half-reactions, we can see that 2 moles of electrons are transferred. Thus, \(n=2\). 2. Determine the reaction quotient (Q).
The expression for the reaction quotient, Q, includes the concentrations of the aqueous species. The activities (concentrations) of pure solids are taken as 1. \[ Q = \frac{[\text{Products}]}{[\text{Reactants}]} = \frac{[Zn^{2+}]}{[Cu^{2+}]} \] Substitute the given concentrations: \[ Q = \frac{0.001 \text{ M}}{0.1 \text{ M}} = \frac{10^{-3}}{10^{-1}} = 10^{-2} \] 3. Apply the Nernst Equation.
We have all the necessary values: - \(E^{\circ}_{cell} = 1.1 \text{ V}\) - \(n = 2\) - \(Q = 10^{-2}\) Substitute these into the Nernst equation: \[ E_{cell} = 1.1 - \frac{0.059}{2}\log_{10}(10^{-2}) \] Since \(\log_{10}(10^{-2}) = -2\), the equation becomes: \[ E_{cell} = 1.1 - \frac{0.059}{2}(-2) \] \[ E_{cell} = 1.1 + 0.059 \] \[ E_{cell} = 1.159 \text{ V} \] Step 4: Final Answer
The cell potential at the given concentrations is 1.159 V.