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A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electr

Marat540 [252]

Complete Question

$Fe^{2+} + 2e^-----> Fe \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = - 0.441 V$

$Cd^{2+} + 2e^- -----> Cd \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^0_{red} = -0.403V$

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03305 V?

Answer:

The mass is M= 117.37g

Explanation:

The overall reaction is as follows

$Cd^{2+} + Fe Fe^{2+} + Cd$

The reaction is this way because the potential of $Cd^{2+} \ reducing \ to \ Cd$ is higher than the potential of $Fe^{2+} \ reducing \ to \ Fe$ so the the Fe would be oxidized and $Cd^{2+}$ would be reduced

At equilibrium the rate constant of the reaction is

$Q = \frac{concentration \ of \ product }{concentration \ of reactant }$

$= \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}$

The Voltage of the cell $E_{cell} = E_{Cd^{2+}/Cd } + E_{Fe^{2+} /Fe}$

Substituting the given values into the equation

$E_{cell} = -0.403 -(-0.441)$

$= 0.038V$

The voltage of the cell at any point can be calculated using the equation

$E = E_{cell} - \frac{0.059}{n_e} Q$

Where $n_e \ is \ the \ number\ of\ electron$

Substituting for Q

$E = E_{cell} - \frac{0.059}{n_e} \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}$

When E = 0.03305 V

$E = E_{cell} - \frac{0.59}{n_e} \frac{Fe^{2+}}{Cd^{2+}}$

Since we are considering the Cd electrode the equation becomes

$E= E_{cell} - \frac{0.059}{n_e} [\frac{1}{Cd^{2+}} ]$

Substituting values and making [ $Cd^{2+}$ ] the subject

$[Cd^{2+}] =\frac{1}{e^{[\frac{0.03305- 0.038}{\frac{0.059 }{2} }] }}$

$= 0.8455M$

Given from the question that the volume is 1 Liter

The number of mole = concentration * volume

= 0.8455 * 1

= 0.8455 moles

At the standard state the concentration of $Cd^{2+}$ is =1 mole /L

Hence the amount deposited on the Cd electrode would be

= Original amount - The calculated amount

= 1 - 0.8455

= 0.1545 moles

The mass deposited is mathematically represented as

$mass = mole * molar \ mass$

The Molar mass of Cd $= 112.41 g/mol$

Mass $= 0.1545 *112.41$

$= 17.37g$

Hence the total mass of the electrode is = standard mass + calculated mass

M= 100+ 17.37

M= 117.37g

6 0

3 years ago

Read 2 more answers

The gravitational potential is a distance z=0.17 m away from a distribution of mass that has the following equation:

sveta [45]

gravitational potential is given as

$V = \frac{GM}{R^2}(R^2 + Z^2)^0.5$

$E = -\frac{dV}{dz}$

$E = -\frac{GM}{R^2}\frac{d}{dz}(R^2+z^2)^0.5$

$E = -\frac{GM}{R^2}*0.5(R^2+z^2)^-0.5*2z$

$E = -\frac{GM*z}{R^2*(R^2 + z^2)^0.5}$

now plug in all values given

M = 110 kg

R = 0.55 m

z = 0.17 m

$E = -\frac{6.67 * 10^{-11}* 110*0.17}{0.55^2*(0.55^2 + 0.17^2)^0.5}$

$E = 7.16 * 10^{-9} N/kg$

so above is the field intensity

4 0

3 years ago

10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)

Darya [45]

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .

Given: Weight = 475 N

$g= 9.8\ m/s^2$

Substitute all values in formula , we get

$475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg$

Hence, his mass = 48.47 kg.

7 0

3 years ago

Question 4 of 10

ch4aika [34]

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

4 0

3 years ago

In gas chromatography, what are the advantages of (a) temperature programming? (check all that apply.)

TiliK225 [7]

The following statements apply:
1. Resolution of low boiling solutes is maintained.
2. Retention times of high boiling solutes are decreased.
Temperature programming refers to the process of increasing the temperature of gas chromatography column as a function of time. Temperature programming is usually applied to samples which contain a mixture of components whose boiling points are within narrow ranges

3 0

3 years ago