Question

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

Answer 1

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