Based on the data given below: E°_Cr₂O₇²⁻/Cr³⁺ = 1.33 V E°_Cl₂/Cl⁻ = 1.36 V$$E^{\circ}_{\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}/\mathrm{Cr}^{3+}} = 1.33\mathrm{ V} \quad E^{\circ}_{\mathrm{Cl}_{2}/\mathrm{Cl}^{-}} = 1.36\mathrm{ V}$$ E°_MnO₄⁻/Mn²⁺ = 1.51 V E°Cr³⁺/Cr = -0.74 V$$E^{\circ}_{\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}} = 1.51\mathrm{ V} \quad E^{\circ}_{\mathrm{Cr}^{3+}/\mathrm{Cr}} = -0.74\mathrm{ V}$$ the strongest reducing agent is :

Core Logic

The disproportionation of Cl₂$Cl_2$ in cold, dilute alkaline medium yields chloride (Cl^-$Cl^-$) and hypochlorite (ClO^-$ClO^-$) ions.

Oxidation half-reaction: Cl₂ arrow 2ClO^- + 2e^-$Cl_2 \rightarrow 2ClO^- + 2e^-$ Balancing O and H in basic medium: Cl₂ + 4OH^- arrow 2ClO^- + 2H₂O + 2e^-$Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-$

Reduction half-reaction: Cl₂ + 2e^- arrow 2Cl^-$Cl_2 + 2e^- \rightarrow 2Cl^-$

Adding both half-reactions (electrons are already equal): 2Cl₂ + 4OH^- arrow 2ClO^- + 2Cl^- + 2H₂O$2Cl_2 + 4OH^- \rightarrow 2ClO^- + 2Cl^- + 2H_2O$

Dividing the entire equation by 2 to get the simplest integer coefficients: 1Cl₂ + 2OH^- arrow 1ClO^- + 1Cl^- + 1H₂O$1Cl_2 + 2OH^- \rightarrow 1ClO^- + 1Cl^- + 1H_2O$

Step 1: Coefficient Matching

By comparing with the given equation: a Cl₂ + b OH^- arrow c ClO^- + d Cl^- + e H₂O$a Cl_2 + b OH^- \rightarrow c ClO^- + d Cl^- + e H_2O$ We get: a = 1$a = 1$ b = 2$b = 2$ c = 1$c = 1$ d = 1$d = 1$

Thus, the values are 1, 2, 1, and 1.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p Block Elements

Core Logic

For a species to undergo disproportionation, it must contain an element that is present in an intermediate oxidation state. This allows it to act both as an oxidizing agent (by being reduced to a lower state) and a reducing agent (by being oxidized to a higher state).

Let's evaluate each species:

• H₂O₂$H_2O_2$: Oxygen is in -1$-1$. Can go to 0$0$ (O₂$O_2$) and -2$-2$ (H₂O$H_2O$). (Yes)
• ClO₃^-$ClO_3^-$: Chlorine is in +5$+5$. Can go to +7$+7$ (ClO₄^-$ClO_4^-$) and -1$-1$ (Cl^-$Cl^-$). (Yes)
• P₄$P_4$: Phosphorus is in 0$0$. Can go to -3$-3$ (PH₃$PH_3$) and +1/+3/+5$+1/+3/+5$. (Yes)
• Cl₂$Cl_2$: Chlorine is in 0$0$. Can go to -1$-1$ (Cl^-$Cl^-$) and +1$+1$ (ClO^-$ClO^-$). (Yes)
• Ag$Ag$: Metal in 0$0$ state. Cannot show negative oxidation state. (No)
• Cu⁺¹$Cu^{+1}$: Copper is in +1$+1$. Can go to 0$0$ (Cu$Cu$) and +2$+2$ (Cu²⁺$Cu^{2+}$). (Yes)
• F₂$F_2$: Fluorine is the most electronegative, only shows 0$0$ and -1$-1$. Cannot be oxidized to a positive state. (No)
• NO₂$NO_2$: Nitrogen is in +4$+4$. Can go to +5$+5$ (HNO₃$HNO_3$) and +3$+3$ (HNO₂$HNO_2$). (Yes)
• K^+$K^+$: Potassium is in its highest oxidation state +1$+1$. Cannot be oxidized further. (No)

Step 1: Final Count

The species that can undergo disproportionation are: H₂O₂$H_2O_2$, ClO₃^-$ClO_3^-$, P₄$P_4$, Cl₂$Cl_2$, Cu⁺¹$Cu^{+1}$, and NO₂$NO_2$. Total count is 6.

Pattern Recognition

Rule out maximum oxidation states (K^+$K^+$), minimum oxidation states, and elements like F₂$F_2$ which never show positive oxidation states.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Core Logic

This is a redox reaction in a slightly basic/neutral medium (as indicated by MnO₂$MnO_2$ product). Separate the reaction into two half-cells and balance using the ion-electron method.

Step 1: Reduction Half-reaction

Balance O by adding H₂O$H_2O$ and H by adding OH^-$OH^-$ (or balance with H^+$H^+$ then convert):

Balance charge by adding e^-$e^-$:

Step 2: Oxidation Half-reaction

Balance I:

Balance charge by adding e^-$e^-$:

Step 3: Combining halves

To cancel electrons, multiply equation (1) by 2 and equation (2) by 3:

Add them together:

Step 4: Conclusion

Comparing this with the given equation:

We see b = 6, c = 4, x = 3, y = 2, z = 8$b = 6, c = 4, x = 3, y = 2, z = 8$. Therefore, the value of z is 8.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Related Formula

Core Logic

From the balanced redox equation in an acidic medium, we can see the exact stoichiometry between permanganate, oxalate, and hydrogen ions. For 2$2$ moles of MnO₄^-$MnO_4^-$, 16$16$ moles of H^+$H^+$ are required.

Step 1: Calculating for 1 mole

For 1$1$ mole of MnO₄^-$MnO_4^-$, the number of moles of H^+$H^+$ ions required is:

Note: The official NTA answer was given as 4 initially, but our experts confirm the standard balanced stoichiometry requires 8 moles of H^+$H^+$. We are outputting the chemically correct value of 8.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The d- and f-Block Elements