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Consider the following ionic equation: Cr\(_2\)O\(_7\)\(^{2-}\) + 14H\(^+\) + ne-  → 2Cr\(^{3+}\) + 7H\(_2\)O. The value...

Chemistry
WAEC 2016

Consider the following ionic equation:
Cr\(_2\)O\(_7\)\(^{2-}\) + 14H\(^+\) + ne-  → 2Cr\(^{3+}\) + 7H\(_2\)O.
The value of n in the equation is_______?

  • A. 7
  • B. 6
  • C. 3
  • D. 2
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Correct Answer: Option B
Explanation

Cr\(_2\)O\(_7\)\(^{2-}\) + 14H\(^+\) + ne-  → 2Cr\(^{3+}\) + 7H\(_2\)O.

All you need to do is to balance the chemical equation, balancing the ionic/ number of moles of the electrons on both sides

Cr\(_2\)O\(_7\)\(^{2-}\) + 14H\(^+\) + ne-  → 2Cr\(^{3+}\) + 7H\(_2\)O.

    (1 x - 2) +  (14 x +1) + n  =  ( 2 x +3)

       - 2  + 14 + n = +6

           12  + n = + 6

What do we need to the +12 on the left hand side to equate +6 on the right hand side

   +12 - 6 = +6

Therefore, number of electrons is 6. i.e 6e\(^-\) 


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