Answer to Question #154712 in Organic Chemistry for 123

Question

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

Solution.

a.

(A): $Zn^{2+} + 2e = Zn^0$

(K): $MnO4^- + 2 H2O + 3e = MnO2 + 4OH^-$

b.

Electrons move from the anode (zinc) to the cathode (platinum).

c.

$Zn(s)|Zn^{2+}(aq)||MnO4^-(aq)|Pt(s)$

d.

$\Delta E = E(k)-E(a)$

$\Delta E = 0.6 - (-0.763) = 1.363 \ V$

Answer:

a.

(A): $Zn^{2+} + 2e = Zn^0$

(K):$MnO4^- + 2 H2O + 3e = MnO2 + 4OH^-$

b.

Electrons move from the anode (zinc) to the cathode (platinum).

c.

Zn(s)|Zn^{2+}(aq)||MnO4^-(aq)|Pt(s)

d.

$\Delta E = 1.363 \ V$