In April 2024
Answer to Question #134827 in General Chemistry for Hugh
4. 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:
a. The half reactions
b. Flow of electrons
c. Short-form notation
d. Electric potential (voltage)
a. Zn+2Cl-=ZnCl2+2e-
Pt(MnO4)2+2e-=Pt2++2MnO4-
b. Zn= Zn2+ + 2e-
Pt= Pt2+ + 2e-
Zn + Pt 2+ → Pt+ Zn2+
c.
Zn(s) ZnCl2 (s) Cl- (aq) Pt(MnO4)2(s) Pt(MnO4)(s) Pt(s)
d. ϕ(Zn)=-0.763 B
ϕ(Pt)=+1.19 B
0.763+1.19=1.953 B
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How is this possible? Isn’t Pt an inert electrode and would not be consumed in the process?
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