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

Question

A chemist designs a galvanic cell that uses these two half-reactions: half-reaction MnO4(aq)+2H2O(,)+3e MnO2(s)+4OH (aq) Br2(l)+ 2e- 2 Br-(aq) standard reduction potential Ered = +0.59 V Ered = + 1.065 V Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions? Yes No If you said it was possible to calculate the cell voltage, do so and enter your answerV here. Round your answer to 2 significant digits

Explanation / Answer

a)

cathode --> Reduction, the highest potential goes here

Br2(l) + 2e- --> 2Br-(aq)

when balanced with the other reaction

3Br2(l) + 6e- --> 6Br-(aq)

b)

anode --> oxidation, lowest potential, invert it

MnO2(s) + 4OH-(aq) --> MnO4-(aQ) + 2H2O8l)+ 3e-

balance

2MnO2(s) + 8OH-(aq) --> 2MnO4-(aq) + 4H2O(l)+ 6e-

c)

now, add all

2MnO2(s) + 8OH-(aq) --> 2MnO4-(aq) + 4H2O(l)+ 6e-

3Br2(l) + 6e- --> 6Br-(aq)

2MnO2(s) + 8OH-(aq) + 3Br2(l) + 6e- --> 6Br-(aq)+2MnO4-(aq) + 4H2O(l)+ 6e-

simplify

2MnO2(s) + 8OH-(aq) + 3Br2(l) --> 6Br-(aq)+2MnO4-(aq) + 4H2O(l)

d)

yes, for STD conditions, we do

e)

E°cell = Ecathode - Eanode = 1.065-0.59 = 0.475 V

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