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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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2 H_ toq 4 .^++2 e^- H_2(~g) Using Nernst equation, E_H^+ m_2^circ=E_H^* H_2^O+0.0591/n log .[H^+]^2/[H_2] -0.3=0+0.0591/c 2 xx 2 log [ddotH^+] -0.3=-0.0591-[-log [H^+] 0.3=0.0591 pH pH^circ=0.3/0.0591=5.076 endaligned
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