---
title: "A student designing an electrochemical method to recover copper from a recycling solution uses the standard reduction half-reaction  \\[ \\text{Cu}^{2+}\\text{(aq)}+2e^-\\rightarrow\\text{Cu(s)} \\qquad E^\\circ=+0.34\\ \\text{V} \\]  To combine the half-reaction with another half-reaction, the student multiplies all its coefficients by \\(2\\).  \\[ 2\\text{Cu}^{2+}\\text{(aq)}+4e^-\\rightarrow2\\text{Cu(s)} \\]  Which statement correctly gives the value of \\(E^\\circ\\) for the multiplied half-reaction and explains why?"
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url: "https://nerd-notes.com/ubq/119960/"
date_modified: "2026-08-21T08:31:05+00:00"
---

# A student designing an electrochemical method to recover copper from a recycling solution uses the standard reduction half-reaction

\[
\text{Cu}^{2+}\text{(aq)}+2e^-\rightarrow\text{Cu(s)} \qquad E^\circ=+0.34\ \text{V}
\]

To combine the half-reaction with another half-reaction, the student multiplies all its coefficients by \(2\).

\[
2\text{Cu}^{2+}\text{(aq)}+4e^-\rightarrow2\text{Cu(s)}
\]

Which statement correctly gives the value of \(E^\circ\) for the multiplied half-reaction and explains why?

A student designing an electrochemical method to recover copper from a recycling solution uses the standard reduction half-reaction

\[
\text{Cu}^{2+}\text{(aq)}+2e^-\rightarrow\text{Cu(s)} \qquad E^\circ=+0.34\ \text{V}
\]

To combine the half-reaction with another half-reaction, the student multiplies all its coefficients by \(2\).

\[
2\text{Cu}^{2+}\text{(aq)}+4e^-\rightarrow2\text{Cu(s)}
\]

Which statement correctly gives the value of \(E^\circ\) for the multiplied half-reaction and explains why?

- **A.** \(E^\circ\) becomes \(+0.68\ \text{V}\), because standard reduction potential is an extensive quantity that doubles when all coefficients are doubled.
- **B.** \(E^\circ\) becomes \(+0.68\ \text{V}\), because \(\Delta G^\circ\) doubles while \(n\) remains \(2\) in \(E^\circ=-\dfrac{\Delta G^\circ}{nF}\).
- **C.** \(E^\circ\) remains \(+0.34\ \text{V}\), because \(\Delta G^\circ\) is unchanged when all coefficients are doubled.
- **D.** \(E^\circ\) remains \(+0.34\ \text{V}\), because both \(\Delta G^\circ\) and \(n\) double, leaving \(E^\circ=-\dfrac{\Delta G^\circ}{nF}\) unchanged.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119960/*