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Svet_ta [14]
3 years ago
13

If E°(red) of a given half-cell is more negative than E°(red) for a standard hydrogen electrode, the half-cell will:

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
7 0

Answer:

Be the anode

Explanation:

The standard hydrogen electrode is regarded as the standard reference electrode and it has been assigned an electrode potential of 0.0V.

If any substance has an electrode potential that is more negative than hydrogen, then that half cell will function as the anode when connected to the standard hydrogen electrode.

Similarly, any substance that has a more positive electrode potential than hydrogen will serve as the cathode when its half cell is connected to the standard hydrogen electrode.

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The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate
vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

            k = Ae^-(Ea/RT)

where,   k = rate constant

              A = pre-exponential factor

          Ea  = activation energy

             R = gas constant

              T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

         

      T1= (244 + 273.15) K = 517.15 K

      T2= (324+ 273.15) K =597.15 K

    K1  = 6.7 ,     K2 = ?

         R = 8.314 J/mol K

     Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

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Answer:

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Explanation:

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I just need points sorry

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Answer:

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