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 standard reduction potential MnO2(s)+4 H + (aq)+2e- Mn2+(aq)+ 2H20(,) red = + 1.23 V Zn2+ (aq)+2e- Zn(s) Ered =-0.763 V Answer the following questions about this cell. Write a balanced equation for the | MnO4(aq) + half-reactin that happens at the cathode ID:1 e- 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 ntaneous as written. Do you have enough information to O Yes I Don't Know Submit

Explanation / Answer

a)

cathode --> reduction, gains of e-, highest E°potential

MnO2(s) + 4H+(aq) + 2e- --> Mn2+(aq) + 2H2O(l)

b)

anode is the oxidation side, it is loss of electrons, so it is the lowest reduction potential

recall that we must invert it to show oxidation

Zn(s) --> Zn2+(aq) + 2e-

c)

balanced

MnO2(s) + 4H+(aq) + 2e- --> Mn2+(aq) + 2H2O(l)

Zn(s) --> Zn2+(aq) + 2e-

add all

Zn(s) + MnO2(s) + 4H+(aq) + 2e- --> Mn2+(aq) + 2H2O(l) + Zn2+(aq) + 2e-

cancel common terms

Zn(s) + MnO2(s) + 4H+(aq) --> Mn2+(aq) + 2H2O(l) + Zn2+(aq)

d)

yes, we hav eneough info

E°cell = Ecahtode - Eanode = 1.23 - (-0.763) = 1.993 V

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