In the given half-reaction Cr_2O_7^{2-} + ZH^ | vedclass

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(D) To balance the half-reaction $Cr_2O_7^{2-} + ZH^{+} + e^- 	o Cr^{3+} + H_2O$:$1$. Balance the $Cr$ atoms: $Cr_2O_7^{2-} + ZH^{+} + e^- 	o 2Cr^{3

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$14$

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(D) To balance the half-reaction $Cr_2O_7^{2-} + ZH^{+} + e^- 	o Cr^{3+} + H_2O$:$1$. Balance the $Cr$ atoms: $Cr_2O_7^{2-} + ZH^{+} + e^- 	o 2Cr^{3+} + H_2O$.$2$. Balance the $O$ atoms by adding $H_2O$: $Cr_2O_7^{2-} + ZH^{+} + e^- 	o 2Cr^{3+} + 7H_2O$.$3$. Balance the $H$ atoms: There are $14$ hydrogen atoms on the right side,so $Z = 14$.$4$. Balance the charge: The left side has $-2 + 14 = +12$ charge (assuming $6$ electrons are added). The right side has $2 	imes (+3) = +6$. To balance,we need $6$ electrons: $Cr_2O_7^{2-} + 14H^{+} + 6e^- 	o 2Cr^{3+} + 7H_2O$. Thus,$Z = 14$.

In the given half-reaction $Cr_2O_7^{2-} + ZH^{+} + e^- \to Cr^{3+} + H_2O$,find the value of $Z$.

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