JEE Mains 1-Feb-2023 Shift 1 Solved Paper

Section: Chemistry

Question : 54 of 90

Marks: +1, -0

At what pH, given half cell MnO4−(0.1M)∣Mn2+ (0.001 M) will have electrode potential of 1.282 V ? _______ (Nearest Integer)
Given EMnO4−∕Mn2+o=1.54V,‌

2.303RT

=0.059V

[1-Feb-2023 Shift 1]

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