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Answer:
answer us A I hope it helps u
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
<u>( I) Bonds are broken, and the reaction is endothermic.</u>
(2) Bonds are broken , and the reaction is exothermic.
(3) Bonds are formed, and the reaction is endothermic.
<span>(4) Bonds are formed, and the reaction is exothermic.</span>
