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). I attached my answer but am unsure. In particular, I don't know what "determine the flow of electrons" means. As well, what elements the salt bridge is made of and what directions it flows, and what solution is in the platnium cell.
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). I attached my answer but am unsure. In particular, I don't know what "determine the flow of electrons" means. As well, what elements the salt bridge is made of and what directions it flows, and what solution is in the platnium 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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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).
I attached my answer but am unsure. In particular, I don't know what "determine the flow of electrons" means. As well, what elements the salt bridge is made of and what directions it flows, and what solution is in the platnium cell.

Transcribed Image Text:1. a) Oxidation: Zn
2+
Zn
+ 2e
(aq)
Reduction: Pt?+
+ 2e
(aq)
-1
electrons
salt bridge
zinc electrode
platinum electrode
2, 2+
2+
Pt
(-)
(+)
cathode
reduction
anode
oxidation
b) The flow of electrons is from the zinc electrode to the platinum electrode through the wire.
c) Zn | Zn || Pt |Pt)
2+
d) E°cell = E°cathode – E°anode
+2e 1→ Pt)
→ Zm«s)
%3!
E°cathode : Pt
E°anode : Zn
2+
+ 1.20
(aq)
2+
-1
+ 2e
- 0.76
(aq)
E°cel = 1.20 –(- 0.76)
= 1.96V
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