Calculate emf of the following cell of `25^@C` :
`Fe|Fe^(2+)(0.001M)||H^+(0.01M)|H_2(g)(1 bar) pt`
`E_((Fe^(2+)//Fe)) = -0.44V`
`E_((H^+//H_2))^@ = 0.00V`

Explanation

Cell reaction
`Fe + 2H^+ to Fe^(2+) + H_2`
`E_("cell")^(@) = E_("cathode")^@ - E_("anode")^@`
` = 0.00 - (0.44)`
`E_("cell")^@ = 0.44 -(0.0591)/2 10 g ([10^(-3)])/([10^(-2)]^(2))`
`= 0.44 - (0.0591)/2 10 g 10`
`= 0.44 - (0.0591)/2`
`E_("cell")^@ = 0.4105V`.