Calculate the cell potential for the following reaction as written at 57 Solutio

Question

Calculate the cell potential for the following reaction as written at 57

Explanation / Answer

you have not given the reduction potential value for Fe2+ and Cr2+ .....i am using the values from wikipedia Eo(Cr2+/Cr) =-0.91 V Eo(Fe2+/Fe) = -0.44 V Cr(s)+Fe^2+(aq) Cr^2+(aq)+ Fe(s) from this reaction we can say that Cr is getting oxidised ( from 0 to +2) and Fe is getting reduced ( +2 to 0 ) so reaction taking place at cathode : Fe2+ + 2e- ---------> Fe and reaction taking place at anode : Cr ------> Cr2+ + 2e- Eo(cell) = Eocathode - Eoanode = -0.44 - (-0.91) = -0.44 + 0.91 = 0.47 V now if the conc. were at 1 M each ....E(cell) would have been 0.47 V but since it is not equal to 1 M so using Nernst equation .... E(cell) = Eo(cell) - RT/nF ln [Cr2+] / [Fe2+] solids and liquids are not written under ln where Eo(cell) = 0.47 V R = 8.314 T = 57 + 273 = 330 K n = no.of electrons in electrode reaction = 2 F = 96500 C [Cr2+] = 0.853 [Fe2+] = 0.016 putting the values.... E(cell) = 0.47 - ( 8.314 X 330)/(2 X 96500) X ln 0.853/0.016 E(cell) = 0.47 - 2743.62 / 193000 X ln 53.313 E(cell) = 0.47 - 0.0284 X 3.976 E(cell) = 0.47 - 0.113 = 0.357 V please check the maths ....i have given the process feel free to ask any question

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