Question

One litre each of a Buffer solution is taken in two beakers connected through a salt bridge. Platinum electrode is immersed in each beaker and electrysis is carried out for `10,000` sec. using a current of `9.65` Ampere. The Buffer contain `(1)/(3)` mole each of `Na_(2)HPO_(4)` and `NaH_(2)PO_(4)` Initially. using the given data answer the following.

Given: ltbr. `pK_(a) (H_(3)PO_(4)) = 4, pK_(a) (H_(2)PO_(4)^(-)) = 7 , pK_(a) (HPO_(4)^(2-)) = 10`
`log 2 = 0.3, log3 = 0.48`
`2.303 (RT)/(F) = 0.06 at 298K`
After electrolysis if `H_(2)` gas is bubbled at 1 bar pressure on each platinum electrode, what is the `EMF` of resulting Galvenic cell.
A. `1.09V`
B. `0.29V`
C. `0.59V`
D. `0.39V`

Answer 1

Correct Answer - C
`EMF = E_("cathode") - E_("Anode")`
`E_(H^(+)//H_(2)) = E_(H^(+)//H_(2))^(@) -(0.0591)/(2)log.(1)/([H^(+)]^(2))`
`E_(H^(+)//H_(2)) =- 0.059 pH`
`EMF =- 0.0591 pH_("cathode") -(-0.0591 pH_("anode"))`
`= (pH_("anode") -pH_("cathode")) 0.06`
`= (11.91 -2.09) 0.06`
`= 0.59`