Chapter 19: Problem 126
An electrochemical cell is made by placing an iron electrode in \(1.00 \mathrm{~L}\) of \(0.15 \mathrm{M} \mathrm{FeSO}_{4}\) solution and a copper electrode in \(1.00 \mathrm{~L}\) of \(0.036 \mathrm{M} \mathrm{CuSO}_{4}\) solution. a. What is the initial voltage of this cell when it is properly constructed? b. Calculate the final concentration of \(\mathrm{Cu}^{2+}\) in this cell if it is allowed to produce an average current of \(1.25\) amp for \(335 \mathrm{~s}\).
Short Answer
Step by step solution
Write Half-Reactions
Calculate Standard Cell Potential
Calculate Initial Cell Potential Using Nernst Equation
Calculate Moles of Electrons Transferred
Calculate Change in Concentration of Cu2+
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nernst equation
Standard reduction potentials
- If \( E^\circ > 0 \), the reaction is spontaneous in the forward direction.
- If \( E^\circ < 0 \), the reaction is non-spontaneous in the forward direction, but spontaneous in the reverse.
Faraday's laws of electrolysis
- First Law: The mass of a substance altered at an electrode during electrolysis is proportional to the total charge passed through the electrolyte.
- Second Law: The masses of different substances altered by the same amount of electricity are proportional to their equivalent weights.
Half-reactions
- Oxidation half-reaction: This involves the loss of electrons. For example, in the electrochemical cell involving iron, the oxidation half-reaction is represented as \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2\text{e}^- \).
- Reduction half-reaction: This involves the gain of electrons. For copper, the reduction half-reaction is \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \).