Calculate e-cell for the reaction?

[匿名]

2017-07-28 1:27 pm

NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l)E∘=0.96V
ClO2(g)+e−→ClO2−(aq)E∘=0.95V
Cu2+(aq)+2e−→Cu(s)E∘=0.34V
2H+(aq)+2e−→H2(g)E∘=0.00V
Pb2+(aq)+2e−→Pb(s)E∘=−0.13V
Fe2+(aq)+2e−→Fe(s)E∘=−0.45V

Use appropriate data to calculate E∘cell for the reaction.
Fe(s)+2H+(aq)→Fe2+(aq)+H2(g)

回答 (2)

不用客氣

2017-07-28 7:29 pm

✔ 最佳答案

Given reduction potential :
2H⁺(aq) + 2e⁻ → 2H₂(g)…… E° = 0.00 V
Fe²⁺(aq) + 2e⁻ → Fe(s) …… E° = -0.45 V

Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g) …… E°(cell) = ? V

====
Method 1 :
E°(cell) of the reaction
= E°(reduction) - E°(oxidation)
= (0.00 V) - (-0.45 V)
= +0.45 V

====
Method 2 :
2H⁺(aq) + 2e⁻ → 2H₂(g)…… E° = 0.00 V
Fe(s) → Fe²⁺(aq) + 2e⁻ …… E° = +0.45 V

Sum the above two half equations.
Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g) …… E°(cell) = 0.00 + 0.45 V = 0.45 V

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[匿名]

2017-07-28 4:26 pm

0.45

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