JEE MAIN - Chemistry (2020 - 8th January Morning Slot - No. 9)
What would be the electrode potential for the given half cell reaction at pH = 5?
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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)
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
1.52TO1.53
Explanation
E = E0 - $${{0.0591} \over 4}\log {\left[ {{H^ + }} \right]^4}$$
$$ \Rightarrow $$ E = 1.23 + 0.0591 × pH
$$ \Rightarrow $$ E = 1.23 + 0.0591 × (5)
$$ \Rightarrow $$ E = 1.52
$$ \Rightarrow $$ E = 1.23 + 0.0591 × pH
$$ \Rightarrow $$ E = 1.23 + 0.0591 × (5)
$$ \Rightarrow $$ E = 1.52
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