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

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