16. Calculate AG° for the following reaction at 315 K. At equilibrium,[H4lOg] = 0.73 M and [104] = 0.08 M.10, (aq) + 2 H20 (1)H;1O8 (aq) Half ReactionEred3 eAl- 1.66 V+2 eBr22 Br+ 1.07 V+2 eCo2+- 0.28 V+Co3 eCr- 0.74 V2 eCu²+Cu+ 0.34 V2 eF22 F+ 2.87 V+2 eFe2+Fe- 0.44 VeFe3+Fe2++ 0.77 V2 eMg?Mg- 2.37 V+2 eNi2+Ni- 0.28 V2 ePb- 0.13 V2 eSn+Sn2*+ 0.15 V2 eZn2+Zn- 0.76 V+++

Given- [H4IO6] = 0.73 M [IO4-] = 0.08 M T = 315 K To find- ∆Go Formula- ∆Go = -RTlnK

Answered: Calculate AG° for the following…

Calculate AG° for the following reaction at 315 K. At equilibrium, [HAIOs] = 0.73 M and [10,] = 0.08 M. 10, (aq) + 2 H20 (1) H;10s (aq)

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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Concept explainers

Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electrolyte, electrodes, and some form of external power source are the main components required for conducting electrolysis. A partition, such as an ion-exchange membrane or a salt bridge, may also be used, but this is optional. These are primarily used to prevent the products from diffusing near the opposing electrode.

Question

16. Calculate AG° for the following reaction at 315 K. At equilibrium,
[H4lOg] = 0.73 M and [104] = 0.08 M.
10, (aq) + 2 H20 (1)
H;1O8 (aq)

Half Reaction
Ered
3 e
Al
- 1.66 V
+
2 e
Br2
2 Br
+ 1.07 V
+
2 e
Co2+
- 0.28 V
+
Co
3 e
Cr
- 0.74 V
2 e
Cu²+
Cu
+ 0.34 V
2 e
F2
2 F
+ 2.87 V
+
2 e
Fe2+
Fe
- 0.44 V
e
Fe3+
Fe2+
+ 0.77 V
2 e
Mg?
Mg
- 2.37 V
+
2 e
Ni2+
Ni
- 0.28 V
2 e
Pb
- 0.13 V
2 e
Sn+
Sn2*
+ 0.15 V
2 e
Zn2+
Zn
- 0.76 V
+
+
+

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