---
title: "A student models the recovery of copper from a battery-recycling solution by passing a constant current through an electrolytic cell. Copper is deposited at the cathode according to the following half-reaction.  \\(\\text{Cu}^{2+}\\text{(aq)}+2\\text{e}^-\\rightarrow\\text{Cu(s)}\\)  The current is \\(1.93\\text{ A}\\) and is applied for \\(1.00\\times10^3\\text{ s}\\). Assume that all the current is used to deposit copper. The Faraday constant is \\(9.65\\times10^4\\text{ C/mol }\\text{e}^-\\), and the molar mass of copper is \\(63.5\\text{ g/mol}\\). What mass of copper is deposited on the cathode?"
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url: "https://nerd-notes.com/ubq/119364/"
date_modified: "2026-08-19T12:40:00+00:00"
---

# A student models the recovery of copper from a battery-recycling solution by passing a constant current through an electrolytic cell. Copper is deposited at the cathode according to the following half-reaction.

\(\text{Cu}^{2+}\text{(aq)}+2\text{e}^-\rightarrow\text{Cu(s)}\)

The current is \(1.93\text{ A}\) and is applied for \(1.00\times10^3\text{ s}\). Assume that all the current is used to deposit copper. The Faraday constant is \(9.65\times10^4\text{ C/mol }\text{e}^-\), and the molar mass of copper is \(63.5\text{ g/mol}\). What mass of copper is deposited on the cathode?

A student models the recovery of copper from a battery-recycling solution by passing a constant current through an electrolytic cell. Copper is deposited at the cathode according to the following half-reaction.

\(\text{Cu}^{2+}\text{(aq)}+2\text{e}^-\rightarrow\text{Cu(s)}\)

The current is \(1.93\text{ A}\) and is applied for \(1.00\times10^3\text{ s}\). Assume that all the current is used to deposit copper. The Faraday constant is \(9.65\times10^4\text{ C/mol }\text{e}^-\), and the molar mass of copper is \(63.5\text{ g/mol}\). What mass of copper is deposited on the cathode?

- **A.** \(0.318\text{ g}\)
- **B.** \(0.635\text{ g}\)
- **C.** \(1.27\text{ g}\)
- **D.** \(2.54\text{ g}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119364/*