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

Question

A chemist designs a galvanic cell that uses these two half-reactions: 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? If you said it was possible to calculate the cell voltage, do so and enter your answer here. Round your answer to 2 significant digits.

Explanation / Answer

Q1.

cathode --> species that will get reduced, i.e. gains electrons, it must be th eone with highest potential:

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

balanced (6 electrons, so factor is 3)

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

Q2.

anode: must be oxidation, so the second equation with lowest potential

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

balance e-, x2

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

invert since it is oxidation

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

Q3.

overall balance:

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

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

add all

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

cancel common terms

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

Q4 and Q5

Yes, at STD

E° = ERed - EOx = 1.065 - 0.59 = 0.475 V

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