Answer to Question #154254 in Organic Chemistry for 123

Question

1, Balance the following reaction in a basic solution:

I ^- + MnO₄^2- → MnO₂ + I₂

2, 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)

Answer 1

1- 4H2O + 6I^- +2MnO₄^- = 3I₂ + 8OH^- +2MnO₄

2- a)- at anode oxidation reaction-

Zn(S) = Zn²⁺ + 2e^-

At cathode Reduction reaction-

Nl⁺²(aq) +2e^_ = Ni(S)

b)- From anode to cathode

c)- Zn(s) I ZnCl₂(aq) II Ni(No₃)₂(aq) Ni(S)

d)-Electric Potential =

$E= E^o - \frac{RT}{2F}ln \frac{[Zn^+2][Ni]}{[Zn][Ni^+2]}$