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Calćulate the pH of the following hal...

Calćulate the pH of the following half cell Pt, `H_2` / `H_2 SO_4` The oxidation potential is +0.3 V .

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The correct Answer is:
5
Reduction potentail `= -0.3V`
`2H^(+) + 2e^(-) rarr H_(2)`
`E_(N^(+)//H_(2)) = E_(H^(+)//H_(2))^(@)- (0.0591)/(n) "log" (1)/([H^(+)]^(2))`
`0.3= 0 - (0.0591)/(2) (-2 log [H^(+)]) = 0.0591pH`
or `pH= (0.3)/(0.0591)= 5.05 = 5`
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