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प्रश्न
Calculate emf of the following cell at 25 °C :
Fe|Fe2+(0.001 M)| |H+(0.01 M)|H2(g) (1 bar)|Pt (s)
E°(Fe2+| Fe)= −0.44 V E°(H+ | H2) = 0.00 V
उत्तर
For the given cell representation, the cell reaction will be Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
The standard emf of the cell will be
`E_(cell)^0 = E^0 H^+ //H_2 -> E^0 Fe^(2+) //Fe`
`=> E_(cell)^0 = 0 - (-0.44) = 0.44V`
The Nernst equation for the cell reaction at 25° C will be
`E_(cell) = E_(cell)^0 - (0.0591)/2 log [Fe^(2+)]/[H^+]^(2+)`
= `0.44 - (0.059)/2 log (0.001)/(0.01)^2`
=0.44-0.02955(log10)
=0.44-0.02955(1)
=0.41045V 0.41V
The Nernst equation for the cell reaction at 25 º C will be
=0.44-0.02955(log10)
=0.44-0.02955(1)
=0.41045V 0.41V
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