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Question

For the given cell:

Cu(s)|Cu2+(C1M)||Cu2+(C2M)|Cu(s)ChangeinGibbsenergy(ΔG)isnegative,if:


A

C2=2C1

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B

C2=C12

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C

C1=2C2

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D

C1=C2

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Solution

The correct option is A

C2=2C1


The explanation for the correct options:

A. The Gibbs free energy will be negative whenC2=2C1

Step1: Given:Cu(s)|Cu2+(C1M)||Cu2+(C2M)|Cu(s) withG=-ve,R=8.303Jmol-1K-1, T=248K,f=96500

Step2: Formula: Ecell=E°cell-2.303RT/nflog[C1/C2]

Step 3:

Anode:Cu(s)+2e-Cu2+(C2M)Cathode:Cu2+(C1M)Cu(s)+2e-Overall:Cu2+(C1M)Cu2+(C2M)

Ecell=E°cell-2.303RT/nflog[C2/C1]Ecell=0-0.0591/2log[C2/C1]Ecell=-0.0591/2log[C2/C1]log[a/b]=loga-logbEcell=-0.0591/2logC2-logC1ForGnegative,theEcellshouldbepositive.So,logC2-logC1willbenegative.

The explanation for the incorrect options:

B.C2=C12

Here C2/C1=2, The concentration of C1is smaller than C2. Due to this, the value of G will become positive.

C.C1=2C2

The concentration of C2is greater than C1. Due to this Ecellwill become negative and the value of G will be positive.

D.C1=C2

If both the concentrations will be equal Ecell=-0.0591/2log[C1/C1], then Ecellwill become negative and the value of G will be positive.

Thus, the correct option is A C1=2C2.


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