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ikadub [295]
2 years ago
14

Using the standard reduction potentials listed in appendix e, calculate the equilibrium constant for the reaction at 298 k: 3 ce

4+ (aq) + bi (s) + h2o (l) → 3 ce3+ (aq) + bio+ (aq) + 2 h+ (aq)
Chemistry
1 answer:
andrezito [222]2 years ago
7 0

Answer:

2.9×10^65

Explanation:

E°cell= E°cathode- E°anode

E°cell= 1.61 V- 0.32 V= 1.29V

The anode is oxidized while the cathode is reduced hence Bi/BiO+ is the anode while Ce4+/Ce3+ is the cathode

Recall that

R=8.314JK-1

F= 96500 C

n=3 moles of electrons

T=298K

E°cell= RT/nF ln K

lnK= E°cell × nF/RT

ln K = 1.29 × 3×96500/8.314×298

ln K= 150.7

K= 2.9×10^65

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Answer: Thus 24.0 g of SO_2 would be needed.

Explanation:

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2 years ago
Consider the following thermochemical equation: C(s) + O2(g) → CO2(g) ΔH = −393 kJ CO(g) + ½O2(g) → CO2(g) ΔH = −294 kJ What is
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Given data

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