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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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The importance of learning is that it helps the individual to acquire the necessary skills through learning and knowledge so that he can achieve his set goals. An important fact about learning is that it is a means to improve knowledge and gain skills that will help in reaching specific goals.

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3 years ago
A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
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Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}

0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

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3 years ago
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Answer:

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4 years ago
A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th
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To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

R_{eq}= \frac{V}{I}

Here,

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While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

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Therefore the equivalent resistance of the light string is 197.4\Omega

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