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Question

Pt(H2)0.1Maceticacid+0.1Msodiumacetate0.1Mformicacid+0.1MsodiumformatePt(H2)at250C?

Acetic acid has Ka=1.8×105 while formic acid has Ka=2.1×104. What would be the magnitude of the emf of the cell?

A
0.032 volt
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B
0.063 volt
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C
0.0456 volt
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D
0.055 volt
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Solution

The correct option is B 0.063 volt
Ans B
For gives equation
K=KA(2 abtic Acid)Ka(formic Acid)
V=1=18×1052.1×104
=0.08571
lnK=FERT
F=RTFlnK
=8314×2981×96500ln(0.08571) volts
Emf of the all =0.063 volts
Magnitude of EMF of the all =|0.063| (moduloud) volts
=0.063 volts
Ans: 0.063 V

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