Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s) EoCell = 0.78 V If [Cu2+] = 0.3 ,what [Fe2+] is needed so that Ecell = 0.76 V?

[匿名]

2018-02-06 12:11 pm

更新1:

he following half-reactions occur during use of the rechargeable nickel-cadmium battery (a galvanic cell): Cd(OH)2(s) + 2e- --> Cd(s) + 2OH-(aq) NiO(OH)(s) + H2O(l) + e- --> Ni(OH)2(s) + OH-(aq) The battery has a potential of 1.35 V under standard conditions with nickel as the cathode. If the standard reduction potential for the cadmium half reaction is -0.860 V, determine the standard potential for the nickel half reaction.

更新2:

Calculate the vapour pressure of a 5% by mass benzoic acid [C7H6O2(aq)] in ethanol solution at 35 C. The vapour pressure of pure ethanol at this temperature is 13.40 kPa. Present your answer in kPa. please help asap

回答 (1)

wanszeto

2018-02-06 2:19 pm

✔ 最佳答案

Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) …. E°(cell) = 0.78 V
If [Cu²⁺] = 0.3, what [Fe²⁺] is needed so that E(cell) = 0.76 V?

Nernst equation: E(cell) = E°(cell) - [RT/(nF)] ln([Fe²⁺]/[Cu²⁺])
0.76 = 0.78 - [8.314 × 298 / (2 × 96500)] ln([Fe²⁺]/0.3)
[8.314 × 298 / (2 × 96500)] ln([Fe²⁺]/0.3) = 0.02
ln([Fe²⁺]) - ln(0.3) = (0.02 × 2 × 96500) / (8.314 × 298)
ln([Fe²⁺]) = [(0.02 × 2 × 96500) / (8.314 × 298)] + ln(0.3)
[Fe²⁺] = e^{[(0.02 × 2 × 96500) / (8.314 × 298)] + ln(0.3)} = 1.4 M (to 2 sig. fig.)

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The following half-reactions occur during use of the rechargeable nickel-cadmium battery (a galvanic cell):
Cd(OH)₂(s) + 2e⁻ → Cd(s) + 2OH⁻(aq) …… E° = E₁
NiO(OH)(s) + H₂O(l) + e⁻ → Ni(OH)₂(s) + OH⁻(aq) …… E° = -0.860 V
The battery has a potential of 1.35 V under standard conditions with nickel as the cathode. If the standard reduction potential for the cadmium half reaction is -0.860 V, determine the standard potential for the nickel half reaction.

E°(cell) = E°(reduction) - E°(oxidation)
1.35 = (-0.860) - E₁
Standard potential for the nickel half reaction, E₁ = -2.21 V

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