You set up a voltaic cell using the two half-cells below. If the [Sn2*] - 1.0 x 10 M and [AI3+ = 1.0M, calculate this non-standard state cell voltage at 298K.A (aq) + 3 e Al(s) E = -1.66 VSn (aq) +2 e Sn(s) E = -0.14 V

Answered: You set up a voltaic cell using the two…

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

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

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Question

You set up a voltaic cell using the two half-cells below. If the [Sn2*] - 1.0 x 10 M and [AI3+ = 1.0
M, calculate this non-standard state cell voltage at 298K.
A (aq) + 3 e Al(s) E = -1.66 V
Sn (aq) +2 e Sn(s) E = -0.14 V

Expert Solution

Given

A voltaic cell whose cell potential is to be determined.

Solution

Based on the standard electrode potential, it can be identified that Sn²⁺/Sn is the cathode, Al³⁺/Al is the anode. Hence, Al is being oxidized and Sn²⁺ is reduced.

$E_{c e l l}^{0}$ = E_R - E_L

= -0.14 V - (-1.66V)

= +1.52 V

Cell potential can be determined according to Nernst equation;

E_cell = $E_{c e l l}^{0}$ + $\frac{0.0591}{n}$ log $\frac{{[A l^{3 +}]}^{2}}{{[S n^{2 +}]}^{3}}$

=+1.52 V + $\frac{0.0591}{6}$ log $\frac{{[1.0 M]}^{2}}{{[1.0 * 10^{- 7} M]}^{3}}$

= +1.52 V + (0.00985)*21

= +1.52 V + 0.20685

= +1.72685 V

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