What are x, y and z respectively in the follo | vedclass

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(A) To balance the redox reaction,we use the ion-electron method.
Oxidation half-reaction: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
Reduction half-reacti

Vedclass · Accepted Answer

$6, 14, 1$

Vedclass · Answer

(A) To balance the redox reaction,we use the ion-electron method.
Oxidation half-reaction: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
Reduction half-reaction: $Cr_2 O_7^{2-} + 14 H^{+} + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2 O$
Multiplying the oxidation half-reaction by $6$ to balance the electrons:
$6 Fe^{2+} \rightarrow 6 Fe^{3+} + 6 e^-$
Adding both half-reactions:
$6 Fe^{2+} + Cr_2 O_7^{2-} + 14 H^{+} \rightarrow 6 Fe^{3+} + 2 Cr^{3+} + 7 H_2 O$
Comparing this with the given equation $x Fe^{2+} + y H^{+} + z Cr_2 O_7^{2-} \rightarrow \frac{y}{2} H_2 O + x Fe^{3+} + 2 z Cr^{3+}$,we get $x = 6, y = 14, z = 1$.

What are $x, y$ and $z$ respectively in the following reaction?
$xFe^{2+}{_{\text{(aq)}}} + yH^{+}{_{\text{(aq)}}} + zCr_2O_7^{2-}{_{\text{(aq)}}} \rightarrow \frac{y}{2}H_2O_{\text{(l)}} + xFe^{3+}{_{\text{(aq)}}} + 2zCr^{3+}{_{\text{(aq)}}}$

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What are $x, y$ and $z$ respectively in the following reaction? $xFe^{2+}{_{\text{(aq)}}} + yH^{+}{_{\text{(aq)}}} + zCr_2O_7^{2-}{_{\text{(aq)}}} \rightarrow \frac{y}{2}H_2O_{\text{(l)}} + xFe^{3+}{_{\text{(aq)}}} + 2zCr^{3+}{_{\text{(aq)}}}$