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I(2)(s) | I^(-) (0.1M) half cell is conn...

`I_(2)(s) | I^(-) (0.1M)` half cell is connected to a `H^(+)(aq) | H_2("1 bar ") | Pt ` half cell and e.m.f is found to be `0.7714 V `. If ` E_(I_2|I^(-))^(@) = 0.535 V ` , find the pH of `H^(+) | H_2` half - cell .

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The correct Answer is:
3
`Pt, (1)/(2) H_2 // H^(oplus) // I_((0.1))^(Ɵ) //(1)/(2) I_2 , Pt , (1)/(2) H_2 + (1)/(2) I_2 to H^(oplus) +I^(Ɵ)`
` E=E^(@) - (0.0591)/(n) log Q , 0.7714 = (0.535 -0) - (0.0591)/(1) log(H^(oplus))(I^(-))`
`-((0.07714 - 0.535)/(0.0591)) log(H^(oplus)) (0.1) , log(H^(oplus))(0.1)=-((0.2364)/(0.0591))=(4)`
` (H^(oplus))(0.1) = 10^(-4) , (H^(oplus)) = (10^(-4))/(0.1) = 10^(-3) , P^(H) =3 `
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