tached my answer. I am unsure of the reduction half of things. I found the reduction equation in the redox table but am unsure. I also don't know how to draw the salt bridge as I don't know which elements make it and which direction they are moving in. If you could draw in the salt bridge flow that would be great
tached my answer. I am unsure of the reduction half of things. I found the reduction equation in the redox table but am unsure. I also don't know how to draw the salt bridge as I don't know which elements make it and which direction they are moving in. If you could draw in the salt bridge flow that would be great
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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1. Build a voltaic cell with one beaker containing potassium permanganate (KMnO4) solution with a platinum electrode and a second beaker containing zinc chloride with a zinc electrode. Determine the half reactions, flow of electrons, short-form notation, and electric potential (voltage). Draw a picutre.
I attached my answer. I am unsure of the reduction half of things. I found the reduction equation in the redox table but am unsure. I also don't know how to draw the salt bridge as I don't know which elements make it and which direction they are moving in. If you could draw in the salt bridge flow that would be great.
Transcribed Image Text:4. a) Oxidation: Zn
→ Zn
(aq)
2+
+ 2e
Reduction: MnO
-1
+1
2+
+ 8H
4 (aq)
(aq)
+ 5e
→ Mn + 4H200
electrons
salt bridge
zine electrode
platinum electrode
Zn 2+
Pt
zine chloride
(-)
potassium permanganate
(+)
anode
oxidation
cathode
reduction
b) The flow of electrons is from the zinc electrode to the platinum electrode through the wire
(anode to cathode).
2+
c) Zn | Zn || KM nO(aq) | P t)
d) E°cell = E°cathode - E°anode
+ 8H +!
4 (aq)
+ 1.51
2+
+ 4H,0
-1
E° cathode : Mn0
+ 5e-→ Mn
(aq)
(aq)
2+
-1
E°anode : Zn
(aq)
+ 2e
> Zn«)
- 0.76
E°cell = 1.51 – (- 0.76)
= 2.27V
Therefore, the electric potential is 2.27V.
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