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title: "A student constructs a galvanic concentration cell at \\(298\\text{ K}\\) by placing a strip of solid silver, \\(\\text{Ag(s)}\\), into each of two beakers. Beaker 1 contains \\(100\\text{ mL}\\) of \\(0.010\\text{ M}\\ \\text{AgNO}_3\\text{(aq)}\\), and Beaker 2 contains \\(100\\text{ mL}\\) of \\(1.0\\text{ M}\\ \\text{AgNO}_3\\text{(aq)}\\). The two half-cells are connected by a salt bridge, and a wire connects the two silver electrodes through a voltmeter. Based on the Nernst equation at \\(298\\text{ K}\\), \\(E_{\\text{cell}} = E^\\circ_{\\text{cell}} – \\dfrac{0.0592\\text{ V}}{n}\\log Q\\), what is the initial cell potential, \\(E_{\\text{cell}}\\)?"
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url: "https://nerd-notes.com/ubq/119976/"
date_modified: "2026-08-21T08:31:25+00:00"
---

# A student constructs a galvanic concentration cell at \(298\text{ K}\) by placing a strip of solid silver, \(\text{Ag(s)}\), into each of two beakers. Beaker 1 contains \(100\text{ mL}\) of \(0.010\text{ M}\ \text{AgNO}_3\text{(aq)}\), and Beaker 2 contains \(100\text{ mL}\) of \(1.0\text{ M}\ \text{AgNO}_3\text{(aq)}\). The two half-cells are connected by a salt bridge, and a wire connects the two silver electrodes through a voltmeter. Based on the Nernst equation at \(298\text{ K}\), \(E_{\text{cell}} = E^\circ_{\text{cell}} – \dfrac{0.0592\text{ V}}{n}\log Q\), what is the initial cell potential, \(E_{\text{cell}}\)?

A student constructs a galvanic concentration cell at \(298\text{ K}\) by placing a strip of solid silver, \(\text{Ag(s)}\), into each of two beakers. Beaker 1 contains \(100\text{ mL}\) of \(0.010\text{ M}\ \text{AgNO}_3\text{(aq)}\), and Beaker 2 contains \(100\text{ mL}\) of \(1.0\text{ M}\ \text{AgNO}_3\text{(aq)}\). The two half-cells are connected by a salt bridge, and a wire connects the two silver electrodes through a voltmeter. Based on the Nernst equation at \(298\text{ K}\), \(E_{\text{cell}} = E^\circ_{\text{cell}} - \dfrac{0.0592\text{ V}}{n}\log Q\), what is the initial cell potential, \(E_{\text{cell}}\)?

- **A.** \(+0.059\text{ V}\)
- **B.** \(+0.12\text{ V}\)
- **C.** \(+0.24\text{ V}\)
- **D.** \(+0.80\text{ V}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119976/*