1.

Derive expression to calculate emf (reduction) of the following half cells at 25^@C H^+||H_2(Pt) (1 atm)

Answer»

Solution :We have the NERNST equation for aA+bB `leftrightarrow -cC+dD`
`E=E^@-(2.303 RT)/(NF) log""{([C]^c [D]^d)/([A]^d [B]^b)}`
Substituting
`R=8.314 JK^-1 mol^-1`
F=96500 COULOMBS and
`T=25+273=298K`, we get
`E=E^@- .0.0591/n log""{([C]^c [D]^d)/([A]^a [B]^b)}`
For the electrode `H^+|H_2(pt)` the half cell reaction is
`H^+ + e leftrightarrow 1/2 H_2(g)` (reduction)
(1 atm)
`therefore E_(H^+) H_2=E_(H^+, H_2)^@-0.0591/2log""([H_2]^(1/2))/([H^+])`
`=0-0.0591log""1/([H^+]) (E_(H^+,H_2)^@=0` volt)
`=0.0591 log[H^+]`


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