Balance the following ionic equation-A Cr2O72- - H- - I- - Cr3- - I2- H2OB Cr2O72- - Fe2- - H- - Cr3- -Fe3- - H2OC MnO4- -SO32- -H- - Mn2- -SO42- - H2OD MnO4- -H- -Br- - Mn2- -Br2- H2O

A) Separating the equations into half reactions Cr_2O_7^2− + H^+ + I^− → Cr^3+ + I_2+ H_2O Oxidation half reaction: I^− → I_2 I=−1 I=0 Reduction half reacti ...

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Byju's Answer

Standard XII

Chemistry

Methods for Ester Synthesis

Balance the f...

Question

Balance the following ionic equation:

A) $C r_{2} O_{7}^{2 -} + H^{+} + I^{-} \to C r^{3 +} + I_{2} + H_{2} O$

B) $C r_{2} O_{7}^{2 -} + F e^{2 +} + H^{+} \to C r^{3 +} + F e^{3 +} + H_{2} O$

C) $M n O_{4}^{-} + S O_{3}^{2 -} + H + \to M n^{2 +} + S O_{4}^{2 -} + H_{2} O$

D) $M n O_{4}^{-} + H^{+} + B r^{-} \to M n^{2 +} + B r_{2} + H_{2} O$

Open in App

Solution

A)
Separating the equations into half reactions
$C r_{2} O_{7}^{2 -} + H^{+} + I^{-} \to C r^{3 +} + I_{2} + H_{2} O$
Oxidation half reaction:
$I^{-} \to I_{2}$
$I = - 1 I = 0$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} \to C r^{3 +}$
$C r = + 6 C r = + 3$

Balancing atoms other than $O$ and $H$
Oxidation half reaction:
$2 I^{-} \to I_{2}$
$I = - 1 I = 0$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} \to 2 C r^{3 +}$
$C r = + 6 C r = + 3$

Balancing oxygen and hydrogen
As reaction is occurring in acidic medium, using $H_{2} O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms.
Oxidation half reaction:
$2 I^{-} \to I_{2}$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} + 14 H^{+} \to 2 C r^{3 +} + 7 H_{2} O$

Balancing the charges
Oxidation half reaction:
$2 I^{-} \to I_{2} + 2 e^{-}$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$

Adding two half reactions by cancellling the electrons involved
Oxidation half reaction: (2I^− → $I_{2} + 2 e^{-}) \times 3 \dots (i)$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O \dots (i i)$
Adding $(i)$ and $(i i)$ , we get;
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$

B)
Separating the equations into half reactions
$C r_{2} O_{7}^{2 -} + F e^{2 +} + H^{+} \to C r^{3 +} + F e^{3 +} + H_{2} O$
Oxidation half reaction:
$F e^{2 +} \to F e^{3 +}$
$F e = + 2 F e = + 3$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} \to C r^{3 +}$
$C r = + 6 C r = + 3$

Balancing atoms other than $O$ and $H$
Oxidation half reaction:
$F e^{2 +} \to F e^{3 +}$
$F e = + 2 F e = + 3$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} \to 2 C r^{3 +}$
$C r = + 6 C r = + 3$

Balancing oxygen and hydrogen
As reaction is occurring in acidic medium, using $H_{2} O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms.
Oxidation half reaction:
$F e^{2 +} \to F e^{3 +}$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} + 14 H^{+} \to 2 C r^{3 +} + 7 H_{2} O$

Balancing the charges
Oxidation half reaction:
$F e^{2 +} \to F e^{3 +} + e^{-}$
Reduction half reaction:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$

Adding two half reactions by cancelling the electrons involved
Oxidation half reaction:
(Fe^{2+} \rightarrow Fe^{3+} +e^−) \times 6\ldots (i)\)
Reduction half reaction:
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O \dots (i i)$
Adding $(i)$ and $(i i)$ , we get;
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 F e^{2 +} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O$

C)
Separating the equations into half reactions
$M n O_{4}^{-} + S O_{3}^{2 -} + H^{+} \to M n^{2 +} + S O_{4}^{2 -} + H_{2} O$
Oxidation half reaction:
$S O_{3}^{2 -} \to S O_{4}^{2 -}$
$S = + 4 S = + 6$
Reduction half reaction:
$M n O_{4}^{-} \to M n^{2 +}$
$M n = + 7 M n = + 2$

Balancing atoms other than $O$ and H
Oxidation half reaction:
$S O_{3}^{2 -} \to S O_{4}^{2 -}$
$S = + 4 S = + 6$
Reduction half reaction:
$M n O_{4}^{-} \to M n^{2 +}$
$M n = + 7 M n = + 2$

Balancing oxygen and hydrogen
As reaction is occurring in acidic medium, using $H_{2} O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms.
Oxidation half reaction:
$S O_{3}^{2 -} + H_{2} O \to S O_{4}^{2 -} + 2 H^{+}$
Reduction half reaction:
$M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$

Balancing the charges
Oxidation half reaction:
$S O_{3}^{2 -} + H_{2} O \to S O_{4}^{2 -} + 2 H^{+} + 2 e^{-}$
Reduction half reaction:
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Adding two half reactions by cancelling the electrons involved
Oxidation half reaction:
$(S O_{3}^{2 -} + H_{2} O \to S O_{4}^{2 -} + 2 H^{+} + 2 e^{-}) \times 5 \dots (i)$
Reduction half reaction:
$(M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O) \times 2 \dots (i i)$
Adding $(i)$ and $(i i)$ , we get;
$5 S O_{3}^{2 -} + 2 M n O_{4}^{-} + 6 H^{+} \to 5 S O_{4}^{2 -} + 2 M n^{2 +} + 3 H_{2} O$

D)
Separating the equations into half reactions
$M n O_{4}^{-} + H^{+} + B r^{-} \to M n^{2 +} + B r_{2} + H_{2} O$
Oxidation half reaction:
$B r^{-} \to B r_{2}$
$B r = - 1 B r = 0$
Reduction half reaction:
$M n O_{4}^{-} \to M n^{2 +}$
$M n = + 7 M n = + 2$

Balancing atoms other than $O$ and $H$
Oxidation half reaction:
$2 B r^{-} \to B r_{2}$
$B r = - 1 B r = 0$
Reduction half reaction:
$M n O_{4}^{-} \to M n^{2 +}$
$M n = + 7 M n = + 2$

Balancing oxygen and hydrogen
As reaction is occurring in acidic medium, using $H_{2} O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms.
Oxidation half reaction:
$2 B r^{-} \to B r_{2}$
Reduction half reaction:
$M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$

Balancing the charges
Oxidation half reaction:
$2 B r^{-} \to B r_{2} + 2 e^{-}$
Reduction half reaction:
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Adding two half reactions by cancelling the electrons involved
Oxidation half reaction:
$(2 B r^{-} \to B r_{2} + 2 e^{-}) \times 5 \dots (i)$
Reduction half reaction:
$(M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O) \times 2 \dots (i i)$
Adding $(i)$ and $(i i)$ , we get;
$10 B r^{-} + 2 M n O_{4}^{-} + 16 H^{+} \to 5 B r_{2} + 2 M n^{2 +} + 8 H_{2} O$

Suggest Corrections

Balance the following ionic equation: A) Cr2O2−7+H++I−→Cr3++I2+H2O B) Cr2O2−7+Fe2++H+→Cr3++Fe3++H2O C) MnO−4+SO2−3+H+→Mn2++SO2−4+H2O D) MnO−4+H++Br−→Mn2++Br2+H2O