Cr _2 O _7^{2-} rightarrow Cr ^{3+}_{aq}
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(D) The balanced redox reaction in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- 	o 2Cr^{3+} + 7H_2O$
$Sn^{2+} 	o Sn^{4+} + 2e^-$
To balance the

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$3$

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(D) The balanced redox reaction in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- 	o 2Cr^{3+} + 7H_2O$
$Sn^{2+} 	o Sn^{4+} + 2e^-$
To balance the electrons,multiply the second equation by $3$:
$3Sn^{2+} 	o 3Sn^{4+} + 6e^-$
Adding the two half-reactions:
$Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ 	o 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
From the stoichiometry,$1 \ mol$ of $Cr_2O_7^{2-}$ reacts with $3 \ mol$ of $Sn^{2+}$.
Therefore,the number of moles of $Sn^{2+}$ oxidized is $3$.

$Cr _2 O _7^{2-} \rightarrow Cr ^{3+}_{aq}$
How many moles of $Sn^{2+}$ will be oxidized by $1 \ mol \ Cr_2O_7^{2-}$ into $Sn^{4+}$?

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$Cr _2 O _7^{2-} \rightarrow Cr ^{3+}_{aq}$ How many moles of $Sn^{2+}$ will be oxidized by $1 \ mol \ Cr_2O_7^{2-}$ into $Sn^{4+}$?