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 .23v rcd Cl,(g)2e 2CI (aq) red Answer the following questions about this cell write a balanced equation for the half-reaction that happons at tho cethole. write h lanced equation for the half-reaction that happerns at the anode write a balanced cquation for the overall reaction that powers the cell. Be sure the reaction is written Do you have ennugh information to calculate the cell Yes oltage under standard conditions? No f yo sa it was possible to calaulata the rell voltage, do su ?d enter your nswer here. Round your significant digits.

Explanation / Answer

The specie with the highest reduction potential will be reduced at the cathode so in this case

at the cathode the reaction will be:

Cl2(g) + 2e- === 2 Cl- (aq) , remember that reduction happens when a species gains electrons

At the anode the oxidation will take place:

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

The overall reaction is:

Mn2+ (aq) + 2H2O(l) + Cl2(g) ====  MnO2 (s) + 4H+ (aq) + 2 Cl- (aq)

You do have enough information

calculate the cell voltage with:

E cell = E cathode - E anode

E cell = 1.359 - 1.23 = 0.129 V

round it to 0.13 V

*hope you find it useful =)

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