Consider the following equations :
2Fe^{2+} + H2O2 - xA + yB
(in basic medium)
2MnO4^- + 6H^+ + 5H2

Q: Why does $\text{H}_2\text{O}_2$ act differently in the two reactions?

$\text{H}_2\text{O}_2$ is a versatile molecule. It can act as an oxidizing agent (getting reduced to $\text{OH}^-$ or $\text{H}_2\text{O}$) or a reducing agent (getting oxidized to $\text{O}_2$) depending on the strength of the other reagent. Since $\text{MnO}_4^-$ is a very strong oxidizing agent, it forces $\text{H}_2\text{O}_2$ to act as a reducing agent.

Q: How do I know what the products are if they aren't given?

For JEE, you must memorize the standard products of common redox reagents. For example, $\text{MnO}_4^-$ in acidic medium always goes to $\text{Mn}^{2+}$, and $\text{Fe}^{2+}$ oxidizes to $\text{Fe}^{3+}$.

Q: Why did we cancel $10\text{H}^+$ in the acidic reaction?

When we added the two half-reactions, we had $16\text{H}^+$ on the reactant side and $10\text{H}^+$ on the product side. To get the net equation, we subtract the smaller amount from both sides, leaving $6\text{H}^+$ on the reactant side, which perfectly matches the problem statement.

Q: Can I balance these using the oxidation number method instead of half-reactions?

Yes, absolutely! Both methods will yield the exact same balanced equation. However, the ion-electron (half-reaction) method is often less prone to errors when dealing with complex ionic species in specific mediums.

Q: What if the $6\text{H}^+$ in the question didn't match my balanced equation?

If your balanced equation has a different number of $\text{H}^+$ than the given equation, it means either you made a mistake in balancing the electrons, or the given equation is a scaled version (e.g., multiplied by a factor). Always double-check your electron cross-multiplication.

Question

Consider the following equations :
$2	ext{Fe}^{2+} + 	ext{H}_2	ext{O}_2 \longrightarrow x	ext{A} + y	ext{B}$
(in basic medium)
$2	ext{MnO}_4^- + 6	ext{H}^+ + 5	ext{H}_2	ext{O}_2 \longrightarrow x'	ext{C} + y'	ext{D} + z'	ext{E}$
(in acidic medium)
The sum of the stoichiometric coefficients $x, y, x', y'$ and $z'$ for products A, B, C, D and E, respectively, is ......... .

Sigma Percentile · Accepted Answer

Balancing redox reactions is a fundamental skill in chemistry, and this problem tests your ability to handle two different environments: basic and acidic mediums. Let's break down the chemistry behind these transformations step-by-step.

### Analyzing the Basic Medium Reaction

We are given the incomplete reaction:
$$2	ext{Fe}^{2+} + 	ext{H}_2	ext{O}_2 \longrightarrow x	ext{A} + y	ext{B}$$

In a basic medium, hydrogen peroxide ($	ext{H}_2	ext{O}_2$) acts as an oxidizing agent. It oxidizes the iron(II) ion ($	ext{Fe}^{2+}$) to iron(III) ion ($	ext{Fe}^{3+}$). The standard half-reactions are:

**Oxidation:**
$$	ext{Fe}^{2+} \longrightarrow 	ext{Fe}^{3+} + e^-$$

**Reduction:**
$$	ext{H}_2	ext{O}_2 + 2e^- \longrightarrow 2	ext{OH}^-$$

To balance the electron transfer, we must multiply the oxidation half-reaction by 2 so that the number of electrons lost equals the number of electrons gained.

Adding the two half-reactions together, the electrons cancel out, yielding the balanced net equation:
$$2	ext{Fe}^{2+} + 	ext{H}_2	ext{O}_2 \longrightarrow 2	ext{Fe}^{3+} + 2	ext{OH}^-$$

Comparing this with the given equation $x	ext{A} + y	ext{B}$, we can clearly identify the products and their stoichiometric coefficients. The products are $2	ext{Fe}^{3+}$ and $2	ext{OH}^-$. Therefore, we find:
$$x = 2$$
$$y = 2$$

### Analyzing the Acidic Medium Reaction

Next, we look at the second incomplete reaction:
$$2	ext{MnO}_4^- + 6	ext{H}^+ + 5	ext{H}_2	ext{O}_2 \longrightarrow x'	ext{C} + y'	ext{D} + z'	ext{E}$$

In an acidic medium, the permanganate ion ($	ext{MnO}_4^-$) is a powerful oxidizing agent. It forces hydrogen peroxide to act as a reducing agent. Permanganate gets reduced to manganese(II) ion ($	ext{Mn}^{2+}$), while hydrogen peroxide gets oxidized to oxygen gas ($	ext{O}_2$). The standard half-reactions are:

**Reduction:**
$$	ext{MnO}_4^- + 8	ext{H}^+ + 5e^- \longrightarrow 	ext{Mn}^{2+} + 4	ext{H}_2	ext{O}$$

**Oxidation:**
$$	ext{H}_2	ext{O}_2 \longrightarrow 	ext{O}_2 + 2	ext{H}^+ + 2e^-$$

To balance the electrons, we find the lowest common multiple of 5 and 2, which is 10. We multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5:

$$2	ext{MnO}_4^- + 16	ext{H}^+ + 10e^- \longrightarrow 2	ext{Mn}^{2+} + 8	ext{H}_2	ext{O}$$
$$5	ext{H}_2	ext{O}_2 \longrightarrow 5	ext{O}_2 + 10	ext{H}^+ + 10e^-$$

Adding these together gives the initial net equation:
$$2	ext{MnO}_4^- + 16	ext{H}^+ + 5	ext{H}_2	ext{O}_2 \longrightarrow 2	ext{Mn}^{2+} + 8	ext{H}_2	ext{O} + 5	ext{O}_2 + 10	ext{H}^+$$

Notice that we have $	ext{H}^+$ ions on both sides. We can simplify the equation by canceling $10	ext{H}^+$ from both sides, leaving exactly $6	ext{H}^+$ on the reactant side:
$$2	ext{MnO}_4^- + 6	ext{H}^+ + 5	ext{H}_2	ext{O}_2 \longrightarrow 2	ext{Mn}^{2+} + 8	ext{H}_2	ext{O} + 5	ext{O}_2$$

This perfectly matches the reactant side provided in the problem statement! Comparing our balanced products with $x'	ext{C} + y'	ext{D} + z'	ext{E}$, we extract the coefficients:
$$x' = 2$$
$$y' = 8$$
$$z' = 5$$

### The Final Calculation

The problem asks for the sum of all these stoichiometric coefficients. We simply add the values we found:
$$	ext{Sum} = x + y + x' + y' + z'$$
$$	ext{Sum} = 2 + 2 + 2 + 8 + 5$$
$$	ext{Sum} = 19$$

**The final answer is 19.** Mastering these standard half-reactions is crucial for quickly solving complex stoichiometry problems in competitive exams.

Enter Numerical Value:

Visualized Solution

$Analyzing the Redox Reactions$

$Half-Reactions in Basic Medium$

$Balancing the Basic Medium Reaction$

$Coefficients x and y$

$Half-Reactions in Acidic Medium$

$Balancing the Acidic Medium Reaction$

$Net Acidic Equation$

$Coefficients x^{'}, y^{'} and z^{'}$

$Sum of All Coefficients$

$Key Takeaways$

The Sigma Insight: Chemical Equation Balancing

Analyzing the Basic Medium Reaction

Analyzing the Acidic Medium Reaction

The Final Calculation

Similar Questions

Consider the following reaction, $x MnO_{4}^{-} + y C_{2} O_{4}^{2 -} + z H^{+} ⟶ x Mn^{2 +} + 2 y CO_{2} + \frac{z}{2} H_{2} O$ The values of $x$ , $y$ and $z$ in the reaction are, respectively

$2 MnO_{4}^{-} + b C_{2} O_{4}^{2 -} + c H^{+} ⟶ x Mn^{2 +} + y CO_{2} + z H_{2} O$ If the above equation is balanced with integer coefficients, the value of $c$ is ......... .

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In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of $CO_{2}$ is

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In dilute aqueous $H_{2} SO_{4}$ , the complex diaquodioxalatoferrate(II) is oxidized by $MnO_{4}^{-}$ . For this reaction, the ratio of the rate of change of $[H^{+}]$ to the rate of change of $[MnO_{4}^{-}]$ is -

The reaction of sulphur in alkaline medium is given below $S_{8} (s) + a O H^{-} (a q) ⟶ b S^{2 -} (a q) + c S_{2} O_{3}^{2 -} (a q) + d H_{2} O (l)$ The value of 'a' is ....... . (Integer answer)

The amount of water produced (in g) in the oxidation of $1$ mole of rhombic sulphur by conc. $HNO_{3}$ to a compound with the highest oxidation state of sulphur is ______. (Given data : Molar mass of water = $18 g mol^{- 1}$ )

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Redox Reactions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Half-Reactions in Basic Medium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing the Basic Medium Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coefficients x and y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Half-Reactions in Acidic Medium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Balancing the Acidic Medium Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Acidic Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coefficients x′,y′ and z′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of All Coefficients

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Key Takeaways