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oee [108]
3 years ago
14

When equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value. 1. false 2. true?

Physics
2 answers:
tigry1 [53]3 years ago
7 0

False

Voltage in an electrochemical cell is in indication of equilibrium, higher will be the non-equilibrium, higher will be the voltage, or we can say at equilibrium voltage tends to 0.

Voltage in an electrical cell is the result of flow of electron, which flow due to difference in charge of the cells, higher the charge difference higher will be the voltage, as the equilibrium between the chemical cells established the flow of electron will stop, and the voltage of the cell tend to 0.

WITCHER [35]3 years ago
7 0

Answer: The given statement is false.

Explanation:

Voltage in an electrochemical cell helps in determining that the system is out of equilibrium.

When a redox reaction approaches towards equilibrium then there occurs flow of electrons with the advancement in chemical reaction. This means as the chemical reaction proceeds towards completion then flow of electrons also takes place.

Hence, cell potential decreases until the reaction reaches at equilibrium where, \Delta G = 0.

Also,  \Delta G = -nFE. So, at equilibrium voltage becomes equal to zero and current also stops.

Thus, we can conclude that the statement when equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value is false.

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2 years ago
Which of the following calls one might hear a refree make?
mestny [16]

Answer: D) All of the above

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
Strike441 [17]

Answer:

E=3.5(8.98*10^{6}x-2.69*10^{15}t)

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)

Explanation:

The electric field equation of a electromagnetic wave is given by:

E=E_{max}(kx-\omega t) (1)

  • E(max) is the maximun value of E, it means the amplitude of the wave.
  • k is the wave number
  • ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

k=\frac{2\pi}{\lambda}            

k=8.98*10^{6} [rad/m]      

And the relation between λ and f is:                

c=\lambda f

f=\frac{c}{\lambda}

f=\frac{3*10^{8}}{700*10^{-9}}

f=4.28*10^{14}

The angular frequency equation is:

\omega=2\pi f

\omega=2\pi*4.28*10^{14}

\omega=2.69*10^{15} [rad/s]

Therefore, the E equation, suing (1), will be:

E=3.5(8.98*10^{6}x-2.69*10^{15}t) (2)

For the magnetic field we have the next equation:

B=B_{max}(kx-\omega t) (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

E_{max}=cB_{max}

B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}

B_{max}=1.17*10^{-8}T

Putting this in (3), finally we will have:

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t) (4)

I hope it helps you!

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4 0
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