Question

Balance the following redox reactions by ion-electron method:
(a)MnO4- (aq) + I - (aq) → MnO2(s) + I2(s) (in basic medium)
(b) MnO4- (aq) + SO2(g) → Mn2+(aq) +HSO4- (aq) (in acidic solution)
(c) H2O2(aq)+Fe2+(aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O72-+ SO2(g) → Cr3+ (aq) +SO42- (aq) (in acidic solution)

Answer 1

(a) Step 1: The two half-reactions involved in the given reaction are:
Oxidation half-reaction: I(aq) $\xrightarrow[]{}$ I2( s)
Reduction half-reaction: MnO4- (aq) $\xrightarrow[]{}$ MnO2(aq)

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Step 2: Balancing I in the oxidation half-reaction, we have:
2I(aq)$\xrightarrow[]{}$I2(s)
Now, to balance the charge, we add 2 e - to the RHS of the reaction.
2I−(aq)$\xrightarrow[]{}$ I2( s)+2e−

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**Step 3: **In the reduction half-reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.
MnO4−+3e−$\xrightarrow[]{}$MnO2(aq)
Now, to balance the charge, we add 4 OH- ions to the RHS of the reaction as the reaction is taking place in a basic medium.
MnO4−(aq)+3e−$\xrightarrow[]{}$MnO2(aq)+4OH−

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Step 4: In this equation, there are 6O atoms on the RHS and 4O atoms on the LHS. Therefore, two water molecules are added to the LHS.
MnO4−(aq)+2H2O$\xrightarrow[]{}$MnO2(aq)+4OH−

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Step 5: Equalizing the number of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we have:
6I−(aq)$\xrightarrow[]{}$3I2( s)+6e−
2MnO4−(aq)+4H2O+6e−→2MnO2(aq)+8OH−(aq)

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Step 6: Adding the two half-reactions, we have the net balanced redox reaction as:
6I−(aq)+2MnO4(aq)+4H2O$\rightarrow$2MnO2(aq)+8OH−(aq)

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(b) Following the steps as in part (a), we have the oxidation half-reaction as: And the reduction half reaction as:
SO2( g)+2H2O(l)$\xrightarrow[]{}$HSO4−(aq)+3H+(aq)
Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2, and then by adding them, we have the net balanced redox reaction as:

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(c) Following the steps as in part (a), we have the oxidation half-reaction as:
The reduction half-reaction as:
Fe2+(aq)$\xrightarrow[]{}$Fe3+(aq)+e−
Multiplying the oxidation half-reaction by 2 and then adding it to the reduction half-reaction, we have the net balanced redox reaction as:
H2O2(aq)+2Fe2+(aq)+2H+(ac)$\xrightarrow[]{}$2Fe(2q)3++2H2O(i)

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(d) Following the steps in part (a), we have the oxidation half-reaction as:
SO2( g)+2H2O(1)$\xrightarrow[]{}$SO24(2q)+4H+(aq)+2e−
And the reduction half reaction as:
Cr2O72−(aq)+14H+(aq)+6e−$\xrightarrow[]{}$Cr3+(aq)+7H2O(i)
Multiplying the oxidation half-reaction by 3 and then adding it to the reduction half-reaction, we have the net balanced redox reaction as:
Cr2O72-(aq)+3SO2( g)+2H+(aq)$\xrightarrow[]{}$2Cr3+(a9)+3SO2−4(aq)+H2O(l)