A chemist designs a galvanic cell that uses these two half-reactions: standard r

Question

A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential Ered = +0.59 V Ered= +0.771 V half-reaction MnO4(ag)+2H,00+3e- MnO2(s)+40H-(ag) Fe3 +(ag)+ e- Fe2+(ag) Answer the following questions about this cell. Write a balanced equation for the half.3- reaction that happens at the cathode. 2+ aq) + e-+ Fe- (aq Write a balanced equationforthe half-| reaction that happens MnO4(aq) + 2H20(l) MnO2(s) + 40H-(aq) + 3e at the anode. Write a balanced equation for the overall reaction that powersthe cell. Be sure the reaction is spontancous as written. | MnO4(aq) + 2H20(1) + Fe3 +(aq) MnO2 (s) + 40H-(aq) + Fe2+(aq) Do you have enoughYes information to calculate the cell voltage under standard conditions? - No If you said it was possible to calculate the cell voltage, do so and enter your 0.18 V answer here. Round your answer to 2 significant digits

Explanation / Answer

a)

cathode --> must be the highest potential, which is that of Fe+3 --> Fe+2

report as

3Fe3+(aq) 3e- = 3Fe2+(aq)

b)

anoce ---> must be with the lower potential, that is, the MnO4-

also, you must invert tis so:

MnO2(s) + 4OH-(aq) = MnO4-(aQ) + 2H2O(l) + 3e-

c)

we do have

d)

E°cell = Ecathode - Eanode = 0.771 -0.59 = 0.181 V

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