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aliya0001 [1]
3 years ago
6

12. Which of the following statements is true about a nickel-cadmium dry cell? (Points : 3) The cathode reaction is Cd + 2OH Cd(

OH)2 + 2e. The cathode reaction is Cd + NiO2 + 2H2O Cd(OH)2 + Ni(OH)2. The cathode reaction is NiO2+ H2O + 2e Ni(OH)2 + 2OH. The anode reaction is NiO2 + H2O + 2e Ni(OH)2 + 2OH.
Chemistry
1 answer:
ella [17]3 years ago
5 0
The correct answer among the options given is the first one. The cathode reaction for a nickel-cadmium dry cell is Cd + 2OH Cd(OH)2 + 2e. It should be remembered that at the cathode reduction always happen and reduction is the gain of electrons. As we can see from the half reaction, there is a gain of 2 electrons.
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Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

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Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

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P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

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P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

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