---
title: "A galvanic cell is constructed at \\(298\\text{ K}\\) according to the balanced chemical equation below. \\[ \\text{Zn}(s) + \\text{Cu}^{2+}(aq) \\rightarrow \\text{Zn}^{2+}(aq) + \\text{Cu}(s) \\quad E^\\circ_{\\text{cell}} = +1.10\\text{ V} \\] The cell initially operates under standard conditions with \\([\\text{Zn}^{2+}] = 1.0\\text{ M}\\) and \\([\\text{Cu}^{2+}] = 1.0\\text{ M}\\). Distilled water is added to the cathode half-cell until the volume of the cathode solution is doubled, while the anode compartment remains unchanged. Which of the following best predicts and explains the effect of this dilution on the cell potential, \\(E_{\\text{cell}}\\), immediately after mixing?"
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url: "https://nerd-notes.com/ubq/120361/"
date_modified: "2026-08-23T04:23:24+00:00"
---

# A galvanic cell is constructed at \(298\text{ K}\) according to the balanced chemical equation below. \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \quad E^\circ_{\text{cell}} = +1.10\text{ V} \] The cell initially operates under standard conditions with \([\text{Zn}^{2+}] = 1.0\text{ M}\) and \([\text{Cu}^{2+}] = 1.0\text{ M}\). Distilled water is added to the cathode half-cell until the volume of the cathode solution is doubled, while the anode compartment remains unchanged. Which of the following best predicts and explains the effect of this dilution on the cell potential, \(E_{\text{cell}}\), immediately after mixing?

A galvanic cell is constructed at \(298\text{ K}\) according to the balanced chemical equation below. \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \quad E^\circ_{\text{cell}} = +1.10\text{ V} \] The cell initially operates under standard conditions with \([\text{Zn}^{2+}] = 1.0\text{ M}\) and \([\text{Cu}^{2+}] = 1.0\text{ M}\). Distilled water is added to the cathode half-cell until the volume of the cathode solution is doubled, while the anode compartment remains unchanged. Which of the following best predicts and explains the effect of this dilution on the cell potential, \(E_{\text{cell}}\), immediately after mixing?

- **A.** \(E_{\text{cell}}\) will increase because decreasing \([\text{Cu}^{2+}]\) causes \(Q < 1\), which increases the thermodynamic driving force for the forward reaction.
- **B.** \(E_{\text{cell}}\) will increase because doubling the solution volume in the cathode half-cell allows a greater total number of \(\text{Cu}^{2+}\) ions to contact the electrode surface.
- **C.** \(E_{\text{cell}}\) will decrease because the decrease in \([\text{Cu}^{2+}]\) reduces the value of the standard cell potential, \(E^\circ_{\text{cell}}\).
- **D.** \(E_{\text{cell}}\) will decrease because decreasing \([\text{Cu}^{2+}]\) causes \(Q > 1\), which brings the system closer to equilibrium and decreases the thermodynamic driving force.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120361/*