2-8 2-8 Electrochemistry: Voltaic Cells on a particle level, describe what is ha

Question

2-8 2-8 Electrochemistry: Voltaic Cells on a particle level, describe what is happening in the first zinc and copper) in n oxidizing acton in number 2, what is oxidized and what is reduced? What is oxidizing agent and what is the reducing agent 3. For the 4, write the cell diagram for all reactions in number 5. What is a salt bridge? What is the purpose of a salt 6. What is the relationship between Gibbs free energy and cell 7. What is the Nernst equation? bridge? l potentia l? If E cell is positive, is the reaction spontaneous? What is the sign on preciuimic 9 ml., of 0050 M KI solution with 3 mL of 0.050 M P(NO) solution will a precipitate form? The Ksp of Pol: is 9.8× 10 8. I WHAT TO DISCUSS IN YOUR CONCLUSION When writing your conclusion for this activity, make sure to consider discussing: the intent of the experiment; one real-world or practical application for this experiment or portion of this experiment (must include references): Your resulting value(s). Is (are) they reasonable? How reliable was the method you used in this experiment? .

Explanation / Answer

Question number 2-4 is connected to question number 1. I am trying to answer by assuming some points

Answer 2:

as we know that a galvanic cell is one which gives electrical energy from a spontaneous redox reaction.

so it consists of half cells, basically few solid metals are used in a particular solution that contains cations fo the metal electrodes. In this half-cell oxidation-reduction reaction taking place.

so, for the example of Zn and Cu

Zn => Zn2+ + 2e

this is an oxidation reaction taking place in one half-cell

now, Cu2+ + 2e => Cu

so this is a reduction reaction taking place in another half-cell

Ans3:

in the Zn-Cu Galvanic cell, copper readily oxidizes zinc. the reduction potential of copper is higher than that of zinc so it gets reduced and oxidizes zinc.  

So, as the reduction potential of copper is higher

Zn ---> Zn2+ + 2e , ZInc got oxidised.

for another half cell reaction

Cu2+ + 2e -----> Cu , Copper get reduced

so the oxidizing agent is copper, and the reducing agent is copper

Answer 4:

Answer5:

Salt bridge, connect the half-cell oxidation-reduction reactions. it helps to maintain the electrical neutrality inside solution circuit. without the salt bridge, the solution in the anode would be positively charged and the solution in the cathode will be negatively charged. it helps to maintain the flow of electrons from oxidation reaction to reduction reaction.

Answer 6

as we know the work done by electrical power per second is given as the product of emf of the cell and total charge.

so W= nFE(Cell)

w= work done,

n= number of charge

F= Faraday's Constant

Ecell = emf of the cell

now Gibbes free energy is the reversible work done

so Delta G = -W

so, deltaG0 = - nFE0cell

so this is the relation between free energy and cell potential.

now if E0cell = a negative quantity so, the value of delta G0 will be positive. as we know for a spontaneous reaction deltaG should be equal or less than zero. so the reaction will no be spontaneous.

Anser 7:

form the above equation

deltaG0= -nFE0cell

now we know that delta G0 = -RTln K

where K = equilibrium constant

R = Universal gas constant

now from the above two equation

-RTlnK = - nFE0cell

E0cell = (RT/nF ) ln K

this equation is known as Nernst Equation

Answer 8

PbI2 ------> Pb2+ + 2I-

[Pb2+] [I-]2 = 9.8* 10-8

as we know the ion product Q if > Ksp then the ionic solid will precipitate.

so,

we have to calculate [Pb2+] in the 3 mL of 0.050 M Pb(NO3) is

moles of Pb2+ = 3 * (0.050 / 1000) = 1.5*10-4

[Pb2+] = {1.5*10-4 / (9+3) } *1000

=1.25* 10-2

similarly [I-] = 9 * {0.050 /(9+3)}

= 3.75*10-2

now [Pb2+] [I-]2 = 17.57 * 10-6

so the ionic product is greater than Ksp so it will precipitate.

thanks

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