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Oxana [17]
3 years ago
13

How to solve for K when given your anode and cathode equations and voltage

Chemistry
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

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3 years ago
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Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g. *
ycow [4]

Answer:

90%

Explanation:

Percentage yield = ?

Theoretical yield = 50g

Actual yield = 45g

To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (45 / 50) × 100

Percentage yield = 0.9 × 100

Percentage yield = 90%

The percentage yield of the substance is 90%

6 0
3 years ago
What is the concentration of a potassium iodate solution after you complete the following porcedure? Pipette 10 mL of a 0.31 M p
zubka84 [21]

<u>Given:</u>

Initial concentration of potassium iodate (KIO3) M1 = 0.31 M

Initial volume of KIO3 (stock solution) V1 = 10 ml

Final volume of KIO3 V2 = 100 ml

<u>To determine:</u>

The final concentration of KIO3 i.e. M2

<u>Explanation:</u>

Use the relation-

M1V1 = M2V2

M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M

Ans: The concentration of KIO3 after dilution is 0.031 M

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D) Calcium

hope this helps :)

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