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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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Answer:

a rotating disk of dust and gases

Explanation:

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6 0
3 years ago
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when the following redoc reaction is balances (assume acidic solution), how many moles of water appear and on what side
PolarNik [594]

8 moles of water on the right side.

An oxidation-reduction or redox reaction is a reaction that involves the transfer of electrons between chemical items (the atoms, ions, or molecules involved in the reaction).

Redox reactions: the burning of fuels, the corrosion of metals, and even the processes of photosynthesis and cellular respiration involve oxidation and reduction.

Step 1:

MnO4- ----> Mn2+

2Cl- ------> Cl2

Step 2:

MnO4- --> Mn2+ + 4H2O

2Cl- -----> Cl2

Step 3:

8H+ + MnO4- ------> Mn2+ + 4H2O

2Cl- ----->Cl2

Step 4:

8H+ + MnO4- +5e- ------>Mn2+ + 4H2O

2Cl- ----> Cl2+ 2e-

Step 5:

16 H+ +2 MnO4- +10Cl- ----->2 Mn2+ + 8H2O+5Cl2

This is the balanced equation in an acidic medium.

That is 8, right side.

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8 0
2 years ago
How many milliliters of a 0.205 M solution of glucose, C6H12O6, are required to obtain 150.1 g of glucose?
jekas [21]

Answer:

4050 mL

Explanation:

Given data:

Mass of glucose = 150.1 g

Molarity of solution = 0.205 M

Volume of solution = ?

Solution:

Molarity  = number of moles of solute / L of solution.

Now we will calculate the moles of sugar first.

Number of moles = mass/ molar mass

Number of moles = 150.1 g/ 180.156 g/mol

Number of moles = 0.83 mol

Now we will determine the volume:

Molarity  = number of moles of solute / L of solution.

0.205 M = 0.83 mol / L of solution.

L of solution = 0.83 mol / 0.205 M

L of solution = 4.05 L

L to mL conversion:

4.05 L × 1000 mL / 1 L = 4050 mL

7 0
4 years ago
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the
NISA [10]

Answer:

Explanation:

Na2CO3+Ca(NO3)2=CaCO3+2NaNO3

nNa2CO3=0.02

nCa(NO3)2=0.02

mCaCO3=0.02*100=2 gram

nNaNo3=0.04

Cm=2/15

6 0
3 years ago
Will ammonium phosphate ionize or dissociate in water
Maslowich
Ionize in water not dissociate
6 0
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