Q3 **Question 3: Electrochemical Cell Potential Calculation****(a) Consider the Following Redox Reactions:**1. \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E^\circ = -0.76 \, \text{V} \)2. \( \text{Cr}^{2+} + 2e^- \rightarrow \text{Cr} \quad E^\circ = -0.91 \, \text{V} \)**Problem Statement:**Assuming there were 1 mole of \( \text{Zn}^{2+} \) and 0.01 moles of \( \text{Cr}^{2+} \) in the reaction above, what will be the \( E_{\text{cell}} \) for the reaction?**Answer Options:**- (a) 0.209 V- (b) 0.150 V- (c) 0.215 V- (d) 0.090 V- (e) 0.0591/2 V**Explanation:**This question requires calculating the cell potential \( E_{\text{cell}} \) using the given standard reduction potentials and concentrations. The Nernst equation might be needed to adjust the standard potential based on the concentrations provided.

Given equations are : Zn2+ + 2e- -------------> Zn , E0 = -0.76 V Cr2+ + 2e- ------------->…

3. (a) Consider Consider the following: Zn²+2e-→ Zn -0.76 V Cr²+2e- Cr EE -0.91 V Assuming that there were 1 moles of Zn² and 0.01 moles of Cr² in the reaction above, what will be the Ece for the reaction? (a) 0.209 V (b) 0.150 V (c) 0.215 V (d) 0.090 V (e) 0.0591/2 V -

3. (a) Consider Consider the following: Zn²+2e-→ Zn -0.76 V Cr²+2e- Cr EE -0.91 V Assuming that there were 1 moles of Zn² and 0.01 moles of Cr² in the reaction above, what will be the Ece for the reaction? (a) 0.209 V (b) 0.150 V (c) 0.215 V (d) 0.090 V (e) 0.0591/2 V -

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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