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 Br2(1)+2e- 2Br (aq) MnO2(s)+4H (aq)+ 2e Mn. (aq)+2H20() Answer the following questions about this cell. standard reduction potential red +1.065 v red = +1.23 V Write a balanced equation for the half-reaction that happens at the cathode 10 Write a balanced equation for the half-reaction that happens at the Write a balanced equation for the overall reaction that powers the In cell. Be sure the reaction is Do you have O Yes information to calculate the cell voltage under No conditions? If you said it was possible to calculate the cell voltage, do so and enter your answer here. Round your answer to2 significant digits.

Explanation / Answer

the electrode with higher standarad reduction potential acts cathode.

the electrode with lower standarad reduction potential acts anode.

so that, half reactions,

at cathode : reduction

MnO2(S) + 4H^+(aq) + 2e-    -----> Mn^2+(aq) + 2H2O(l)

at anode : oxidation

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

cell reaction : MnO2(S) + 4H^+(aq) + Br2(l) -----> Mn^2+(aq) + 2H2O(l) + 2 Br^-(aq)

yes,it is possible to calculate E0cell

E0cell = E0cathode - E0anode

        = 1.23 - 1.065

        = 0.16 v

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