Answer the following questions about this cell. Write a balanced equation for th

Question

Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. Cl_2(g) + 2e^- Rightarrow 2Cl^-(aq) Write a balanced equation for the half-reaction that happens at the a node. H 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

Explanation / Answer

ANODE : 2H2(g) + 4OH- 4H2O(l) + 4e-

CATHODe: O2(g) + 2H2O(l) + 4e- 4OH-

Overall : 2H2(g) + O2(g) 2H2O(l)

Cl2(g) + 2e- 2Cl- Eo = +1.36 V

O2(g) + 4H+ + 4e- 2H2O(l) Eo = +1.23 V

Oxidizing (Top) Cl2 > O2 > Fe3+ > Fe2+ (Bottom)

Reducing (Bottom) Fe > Fe2+ > H2O > Cl- (Top)

Cl2/Cl- has higher Eo (Cl2/Cl- is above O2,H+/H2O)

Cl2 is a stronger oxidizing agent than O2

Cl2 can oxidize H2O to O2 at standard conditions

Cl2/Cl- has the higher reduction potential (Eo)

Cl2/Cl- undergoes reduction

Cl2 + H2O = HOCl + H+ + Cl-

Cl2 + 2e- = 2Cl- (reduction of Cl)

Cl2 + 2H2O = 2 HOCl + 2H+ + 2e- (oxidation of Cl)

2Cl2 + 2H2O = 2 HOCl + 2H+ + 2 Cl

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