In the half-reaction $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,the equivalent weight of $Cr_2O_7^{2-}$ is obtained by dividing its molecular weight by:

  • A
    $1$
  • B
    $3$
  • C
    $6$
  • D
    $14$

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