---
title: "A galvanic cell is constructed under standard conditions at \\(298\\text{ K}\\) using a nickel electrode immersed in a \\(1.0\\text{ M }\\text{Ni(NO}_3)_2(aq)\\) solution and a silver electrode immersed in a \\(1.0\\text{ M }\\text{AgNO}_3(aq)\\) solution, connected by a salt bridge and an external circuit. The overall cell reaction is represented by the following equation.  \\[ \\text{Ni}(s) + 2\\text{Ag}^+(aq) \\rightleftharpoons \\text{Ni}^{2+}(aq) + 2\\text{Ag}(s) \\quad E^\\circ_{\\text{cell}} = +1.06\\text{ V} \\]  As the cell operates and current flows through the circuit, the measured cell potential \\(E_{\\text{cell}}\\) decreases over time until the voltmeter reads \\(0.00\\text{ V}\\). Which of the following statements best explains why the cell potential reaches \\(0.00\\text{ V}\\)?"
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url: "https://nerd-notes.com/ubq/119925/"
date_modified: "2026-08-21T08:18:14+00:00"
---

# A galvanic cell is constructed under standard conditions at \(298\text{ K}\) using a nickel electrode immersed in a \(1.0\text{ M }\text{Ni(NO}_3)_2(aq)\) solution and a silver electrode immersed in a \(1.0\text{ M }\text{AgNO}_3(aq)\) solution, connected by a salt bridge and an external circuit. The overall cell reaction is represented by the following equation.

\[ \text{Ni}(s) + 2\text{Ag}^+(aq) \rightleftharpoons \text{Ni}^{2+}(aq) + 2\text{Ag}(s) \quad E^\circ_{\text{cell}} = +1.06\text{ V} \]

As the cell operates and current flows through the circuit, the measured cell potential \(E_{\text{cell}}\) decreases over time until the voltmeter reads \(0.00\text{ V}\). Which of the following statements best explains why the cell potential reaches \(0.00\text{ V}\)?

A galvanic cell is constructed under standard conditions at \(298\text{ K}\) using a nickel electrode immersed in a \(1.0\text{ M }\text{Ni(NO}_3)_2(aq)\) solution and a silver electrode immersed in a \(1.0\text{ M }\text{AgNO}_3(aq)\) solution, connected by a salt bridge and an external circuit. The overall cell reaction is represented by the following equation.

\[ \text{Ni}(s) + 2\text{Ag}^+(aq) \rightleftharpoons \text{Ni}^{2+}(aq) + 2\text{Ag}(s) \quad E^\circ_{\text{cell}} = +1.06\text{ V} \]

As the cell operates and current flows through the circuit, the measured cell potential \(E_{\text{cell}}\) decreases over time until the voltmeter reads \(0.00\text{ V}\). Which of the following statements best explains why the cell potential reaches \(0.00\text{ V}\)?

- **A.** \(E^\circ_{\text{cell}}\) decreases to \(0.00\text{ V}\) because the concentrations of \(\text{Ag}^+(aq)\) and \(\text{Ni}^{2+}(aq)\) become equal, causing the reaction quotient \(Q\) to equal \(1\).
- **B.** \(E_{\text{cell}}\) decreases to \(0.00\text{ V}\) because the standard Gibbs free energy change \(\Delta G^\circ\) becomes \(0\text{ kJ/mol}_{\text{rxn}}\) when the limiting reactant is completely consumed.
- **C.** \(E^\circ_{\text{cell}}\) decreases to \(0.00\text{ V}\) because the forward and reverse reaction rates become equal, resulting in a system with \(\Delta G = 0\text{ kJ/mol}_{\text{rxn}}\).
- **D.** \(E_{\text{cell}}\) decreases to \(0.00\text{ V}\) because the reaction quotient \(Q\) increases until it equals \(K\), resulting in a system with \(\Delta G = 0\text{ kJ/mol}_{\text{rxn}}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119925/*