Explain the balancing of a redox reaction by the half-reaction method with a suitable example.
Step-$1$: Write the unbalanced equation for the reaction in ionic form:
$Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Fe_{(aq)}^{+3} + Cr_{(aq)}^{+3}$
Step-$2$: Separate the equation into two half-reactions.
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Cr_{(aq)}^{+3}$
Step-$3$: Balance the atoms other than $O$ and $H$ in each half-reaction. For the reduction half-reaction,multiply $Cr^{+3}$ by $2$ to balance $Cr$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow 2Cr_{(aq)}^{+3}$
Step-$4$: For reactions in acidic medium,add $H_{2}O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$5$: Add electrons to balance the charges. Multiply the oxidation half-reaction by $6$ to equalize the electrons:
Oxidation half: $6Fe_{(aq)}^{+2} \rightarrow 6Fe_{(aq)}^{+3} + 6e^{-}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} + 6e^{-} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$6$: Add the two half-reactions to obtain the balanced overall reaction:
$6Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 6Fe_{(aq)}^{+3} + 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
$Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Fe_{(aq)}^{+3} + Cr_{(aq)}^{+3}$
Step-$2$: Separate the equation into two half-reactions.
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Cr_{(aq)}^{+3}$
Step-$3$: Balance the atoms other than $O$ and $H$ in each half-reaction. For the reduction half-reaction,multiply $Cr^{+3}$ by $2$ to balance $Cr$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow 2Cr_{(aq)}^{+3}$
Step-$4$: For reactions in acidic medium,add $H_{2}O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$5$: Add electrons to balance the charges. Multiply the oxidation half-reaction by $6$ to equalize the electrons:
Oxidation half: $6Fe_{(aq)}^{+2} \rightarrow 6Fe_{(aq)}^{+3} + 6e^{-}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} + 6e^{-} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$6$: Add the two half-reactions to obtain the balanced overall reaction:
$6Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 6Fe_{(aq)}^{+3} + 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Explore More
Similar Questions
The number of moles of acidified $KMnO_4$ required to convert one mole of sulphite ion into sulphate ion is
Medium
View SolutionBalance the following ionic equations:
$1. Cr_{2}O_{7}^{2-} + H^{+} + I^{-} \to Cr^{3+} + I_{2} + H_{2}O$
$2. Cr_{2}O_{7}^{2-} + Fe^{2+} + H^{+} \to Cr^{3+} + Fe^{3+} + H_{2}O$
$3. MnO_{4}^{-} + SO_{3}^{2-} + H^{+} \to Mn^{2+} + SO_{4}^{2-} + H_{2}O$
$4. MnO_{4}^{-} + H^{+} + Br^{-} \to Mn^{2+} + Br_{2} + H_{2}O$
$1. Cr_{2}O_{7}^{2-} + H^{+} + I^{-} \to Cr^{3+} + I_{2} + H_{2}O$
$2. Cr_{2}O_{7}^{2-} + Fe^{2+} + H^{+} \to Cr^{3+} + Fe^{3+} + H_{2}O$
$3. MnO_{4}^{-} + SO_{3}^{2-} + H^{+} \to Mn^{2+} + SO_{4}^{2-} + H_{2}O$
$4. MnO_{4}^{-} + H^{+} + Br^{-} \to Mn^{2+} + Br_{2} + H_{2}O$
Difficult
View SolutionIn an alkaline solution,$ClO_2$ oxidizes $H_2O_2$ to $O_2$ and is itself reduced to $Cl^-$. How many moles of $H_2O_2$ will be oxidized by $1 \ mol$ of $ClO_2$?
Difficult
View SolutionIn alkaline condition $KMnO_4$ reacts as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
Therefore,its equivalent weight will be:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
Therefore,its equivalent weight will be:
Medium
View SolutionAssign $A, B, C, D$ from the given type of reaction.
$Na_2CO_3 + PbSO_4 \longrightarrow PbCO_3 \downarrow + Na_2SO_4$
$Na_2CO_3 + PbSO_4 \longrightarrow PbCO_3 \downarrow + Na_2SO_4$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo