Chapter 9: Problem 81
The emf of a Daniell cell at \(298 \mathrm{~K}\) is \(E_{\mathrm{i}}\)
\(\mathrm{Zn}\left|\mathrm{ZnSO}_{4} \| \mathrm{CuSO}_{4}\right| \mathrm{Cu}\)
\((0.01 \mathrm{M}) \quad(1.0 \mathrm{M})\)
when the concentration of \(\mathrm{ZnSO}_{4}\) is \(1.0 \mathrm{M}\) and that of
\(\mathrm{CuSO}_{4}\) is \(0.01 \mathrm{M}\), the emf changed to \(E_{2}\). What is
the relationship between \(E_{1}\) and \(E_{2} ?\)
(a) \(E_{1}=E_{2}\)
(b) \(E_{2} \neq E_{1}\)
(c) \(E_{1}>E_{2}\)
(d) \(E_{1}
Short Answer
Step by step solution
Understand the Daniell cell
Recall the Nernst Equation
Calculate for initial conditions
Calculate for changed conditions
Compare E1 and E2
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst equation
Electromotive force (emf)
Reaction quotient
Electrochemical cell
- **Anode Reaction**: Zinc loses electrons (oxidation), forming \( \text{Zn}^{2+} \) ions.
- **Cathode Reaction**: Copper ions gain electrons (reduction), depositing \( \text{Cu} \) metal.