6 times 10^{-3} mole K_2Cr_2O_7 reacts comple | vedclass

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(A) In the reaction,$K_2Cr_2O_7$ acts as an oxidizing agent and $X^{n+}$ acts as a reducing agent.The change in oxidation state of $Cr$ in $K_2Cr_2O_7

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$1$

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(A) In the reaction,$K_2Cr_2O_7$ acts as an oxidizing agent and $X^{n+}$ acts as a reducing agent.The change in oxidation state of $Cr$ in $K_2Cr_2O_7$ is from $+6$ to $+3$. The change per $Cr$ atom is $3$,and since there are $2$ $Cr$ atoms,the total change is $2 	imes 3 = 6$.Total electrons gained by $K_2Cr_2O_7 = 6 	imes 10^{-3} 	imes 6 = 36 	imes 10^{-3} 	ext{ mol}$.In $XO_3^-$,the oxidation state of $X$ is $+5$. The change in oxidation state of $X$ is $(5 - n)$.Total electrons lost by $X^{n+} = 9 	imes 10^{-3} 	imes (5 - n)$.Equating electrons gained and lost: $36 	imes 10^{-3} = 9 	imes 10^{-3} 	imes (5 - n)$.$4 = 5 - n$,which gives $n = 1$.

$6 \times 10^{-3}$ mole $K_2Cr_2O_7$ reacts completely with $9 \times 10^{-3}$ mole $X^{n+}$ to give $XO_3^-$ and $Cr^{3+}$. The value of $n$ is

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