Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following occurs at the cathode? (A) \(\mathrm{Cr}^{3+}\) is reduced to \(\mathrm{Cr}(\mathrm{s})\) (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) (C) \(\mathrm{Cr}(s)\) is oxidized to \(\mathrm{Cr}^{3+}\) (D) \(\quad \mathrm{Pb}(s)\) is oxidized to \(\mathrm{Pb}^{2+}\)

Step by step solution

01

Identify the given reduction half-reactions

Two reduction half-reactions are given in this problem: \(Cr^{3+}+3e^- \rightarrow Cr(s)\) with \(E^° = -0.41V\) and \(Pb^{2+} + 2e^- \rightarrow Pb(s)\) with \(E^° = -0.12V\). These half-reactions are written in the reduction form, which means electrons are on the left side.

02

Determine the half-reaction that is more likely to occur

In electrochemical cells, the cathode is where the reduction occurs. In this case, the reduction process with the higher standard potential (more positive value) will occur. \nPb^2+ to Pb has higher standard potential (\(E^°\)) than \( Cr^3+ \) to Cr. Thus, the reaction where \(Pb^{2+}\) is reduced to \(Pb(s)\) will occur at the cathode.

03

Select the correct answer

From the options given, only (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) accurately describes what happens at the cathode. Thus, (B) is the correct answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!