What would be the electrode potential for the given half cell reaction at pH = 5?
______.
2H2O $\to$ O2 + 4H$\oplus$ + 4e– ;
$E_{red}^0$ = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
Answer (integer)
1
Solution
E = E<sub>0</sub> - ${{0.0591} \over 4}\log {\left[ {{H^ + }} \right]^4}$
<br><br>$\Rightarrow$ E = 1.23 + 0.0591 × pH
<br><br>$\Rightarrow$ E = 1.23 + 0.0591 × (5)
<br><br>$\Rightarrow$ E = 1.52
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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