Asked by bekah
Use the following pair of reduction half-reactions to design a galvanic cell. Then write in the proper coefficient for each of the species involved in the overall reaction. Water molecules and protons are not shown in the half-reactions, but may be needed in the overall reaction.
MnO4- (aq)->MnO2 (s)
ξo = 0.59 V
NO3- (aq)->NO (g)
ξo = 0.96 V
NO3- (aq) =
MnO4- (aq) =
NO (g) =
MnO2 (s) =
H+ (aq) =
H2O (l) =
MnO4- (aq)->MnO2 (s)
ξo = 0.59 V
NO3- (aq)->NO (g)
ξo = 0.96 V
NO3- (aq) =
MnO4- (aq) =
NO (g) =
MnO2 (s) =
H+ (aq) =
H2O (l) =
Answers
Answered by
DrBob222
You know we can't draw pictures on this forum but the half reactions are
MnO4^- + 3e + 2H2O ==>MnO2 + 4OH^-
NO + 4OH^- ==> NO3^- + 3e + 2H2O
I have written the MnO4^- as a reduction an the NO as an oxidation so as to keep the electrons on opposite sides. I think you will need to reverse these half reactions in order to make a galvanic cell that produces a + voltage.
MnO4^- + 3e + 2H2O ==>MnO2 + 4OH^-
NO + 4OH^- ==> NO3^- + 3e + 2H2O
I have written the MnO4^- as a reduction an the NO as an oxidation so as to keep the electrons on opposite sides. I think you will need to reverse these half reactions in order to make a galvanic cell that produces a + voltage.
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