JEE Main 2020 — Electrochemistry Question with Solution
From: JEE Main 2020 (Online) 8th January Morning Slot
Question
What would be the electrode potential for the given half cell reaction at pH = 5?
______.
2H2O O2 + 4H + 4e– ; = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
2H2O O2 + 4H + 4e– ; = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
Enter your answer
Show full solutionCorrect answer: 1.52
Correct answer
1.52
Step-by-step explanation
E = E0 -
E = 1.23 + 0.0591 × pH
E = 1.23 + 0.0591 × (5)
E = 1.52
E = 1.23 + 0.0591 × pH
E = 1.23 + 0.0591 × (5)
E = 1.52
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