In the previous post, we have discussed on a MCQ question on Electrolysis that requires students to:
1. Write out discharge half equations
2. Do Molar Ratio Comparisons
Question:
Which of the following requires the least number of eletrons for discharge?
A. 2 mol of Al3+ (aluminium) ions
B. 5 mole of OH- (hydroxides) ions
C. 3 mol of O2- (oxides) ions
D. 6 mol of H+ (hydrogen) ions
Write your Answer and Suggested Working in the “Leave A Reply” section below.
PS: If you are not sure how to proceed, you can refer to the previous blogpost by clicking HERE
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i tink it is b
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my ans is also b
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It’s B.
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Al(3+) + 3e- ——> Al
2mol 6mol
4OH- ——> 2H2O + O2 + 4e-
5mol 5mol
2O(2-)—> O2 + 4e-
3mol 6mol
2H(+) + 2e- —-> H2
6mol 6mol
so answer is B.
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dannie Reply:
November 12th, 2011 at 1:47 pm
see which one needds lowest no of mol of e-
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its b
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i think is d
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Correct answer is B.
A: require 6 electrons
B: require 5 electrons
C&D : require 6 electrons
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Its D, because Option B and C will release electrons to be discharged and only A and D requires e.. D needs 2e Al need 3..
So its D ..:)
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