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Chandan panigrahi Asked a Question
October 1, 2020 9:34 pmpts 30 pts
Ex. The EMF of the cell, Pb | PbCl, || AgCI | Ag at 300 K is 0.50 V. If the temperature EMF is-2 x 104 volt deg-1, calculate AH and AS for the cell reaction, ppp coetficien ec Pb + 2AgCl PbC, + 2Ag Pb PbCL|CH | AgCl | Ag iff- Sol. The cell reaction is Pb + 2AgCl > PbCl, + 2Ag AG =- nFE = -2 x 96500 x 0.5 J =-96,500 J or -96.5 kJ AS =n FE OT = 2x 96500 (-2 x 104) =-38.6 JK1 AG AH-TAS, -96,500 AH 300 (-38.6) AH -96,500 300x 38.6 = -108,080 J or 108.08 kJ sir in this question how it takes E=0.5j as it is given that E=0.50v
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  • Priyanshu kumar thankyou
    electromotive force is not actually a force. It is commonly measured in units of volts, equivalent in the metre–kilogram–second system to one joule per coulomb of electric charge.
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    Priyanshu kumar
    So here it is taken
  • Dinesh khalmaniya
    what's your doubt?
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    Chandan panigrahi
    here it takes E=0.5j but in the question its given that E=0.50v
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