1.

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

Answer»

SOLUTION :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`.


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