1.

Calculate the reduction potential of the following electrode at 298 K : Pt Cl_(2) (2.5 atm) HCl (0.01 M), E^(Theta) Cl_(2) | 2 Cl^(-) = 1.36 V .

Answer»

Solution :The electrode REACTION is :
`Cl_(2) (g) + 2E^(-) to 2 Cl^(-) (aq)`
`E (Cl_(2)|2Cl^(-)) = E^(THETA) (Cl|2Cl^(-)) - ((0.059))/(2) "log" ([Cl^(-)]^(2))/(p (Cl_(2)))`
`p(Cl_(2)) = 2.5` atm , `[Cl^(-)] = 0.1 M` (same as the CONCENTRATION of HCl)
`E^(Theta) (Cl_(2) | 2Cl^(-)) = 1.36` V
`THEREFORE E(Cl_(2) |Cl^(-)) = 1.36 - ((0.059))/(2) "log" ((0.1)^(2))/((2.5))`
`= 1.36 + 0.0295 xx 4.398 `
`= 1.36 + 0.13 = 1.49` V


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