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Value of \(E_{H_2O/H_2}\)(1atm)Pt at 298 K would be |
–0.207 V +0.207 V –0.414 V +0.414 V |
–0.414 V |
For water at 298K, \([H^+] = 10^{-7}M\) The reduction reaction is \[H^+ + e^- \rightarrow \frac{1}{2}H_2\] ∴ \(E = -\frac{RT}{F}ln\frac{P_H^{1/2}}{[H^+]} = -0.0591 log\frac{P_H^{1/2}}{[H^+]}\) \(E =-0.0591 log\frac{1}{10^{-7}}\) \(E = -0.4137 ≈ -0.414 V\) |
