**Electrochemical Cell Analysis at 298 K**Consider a cell undergoing the reaction below at 298 K:\[ 2 \text{Na (s)} + \text{Cd}^{2+} \text{(aq)} \leftrightarrows 2 \text{Na}^{+} \text{(aq)} + \text{Cd (s)} \]When \([ \text{Na}^{+} ] = 0.0005 \text{ molal} \, (\gamma = 0.974)\) and \([ \text{Cd}^{2+} ] = 0.2 \text{ molal} \, (\gamma = 0.102)\), then \(\text{E}_{\text{cell}} = +2.45 \text{ V}\). What is \(\text{E}^\circ\) for the cell?**Explanation**- "Na" represents sodium, which in its solid form is a reactant.- "Cd" is cadmium, with its ions present in aqueous solution in this reaction.- \(\gamma\) represents the activity coefficient for each ion concentration.- \(\text{E}_{\text{cell}}\) denotes the electromotive force (EMF) under the given conditions.- The goal is to find \(\text{E}^\circ\), the standard cell potential.

Answered: Image /qna-images/answer/819cfc4c-e6e2-4211-9ab2-ce905feca11a.jpg

Consider a cell undergoing the reaction below at 298 K 2 Na (s) + Cd²+(aq) <=> 2 Na+ (aq) + Cd (s) When [Na] 0.0005 molal (y=0.974) and [Cd²+] = 0.2 molal (y 0.102), then Ecell =+2.45 V. What is Eº for the cell =

Consider a cell undergoing the reaction below at 298 K 2 Na (s) + Cd²+(aq) <=> 2 Na+ (aq) + Cd (s) When [Na] 0.0005 molal (y=0.974) and [Cd²+] = 0.2 molal (y 0.102), then Ecell =+2.45 V. What is Eº for the cell =

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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