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madreJ [45]
4 years ago
10

A galvanic cell is powered by the following redox reaction:2Fe+3(aq) + H2 (g) + 2OH−(aq) → 2Fe+2 (aq) + 2H2O (l)Answer the follo

wing questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab.Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. =E0V
Chemistry
1 answer:
ICE Princess25 [194]4 years ago
7 0

Answer:

Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻

Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)

E° = 1.60 V

Explanation:

Let's consider the reaction taking place in a galvanic cell.

2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)

The corresponding half-reactions are:

Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻   E°red = - 0.83 V

Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)        E°red = 0.77 V

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V

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Answer:

Explanation:

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