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in Chemistry by (71.4k points)
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Calculate emf of the following cell 

Cd/Cd2+ (.10 M)//H+ (.20 M)/H2 (0.5 atm)/Pt 

[Given E° for Cd2+ /Cd = -0.403 v]

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Ecell = E°cell – 0.0591/n Log [Cd2+]/ [H+]2 

 E°cell = 0 – (–.403 V) =0.403 V 

= 0.0403 – 0.0591/2 Log (0.10) X 0.5/(0.2)2 = 0.400 V 

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