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Cr(2)O(7)^(2-)+H^(+)+SO(3)^(2-) to Cr^(3...

`Cr_(2)O_(7)^(2-)+H^(+)+SO_(3)^(2-) to Cr^(3+)(aq.)+SO_(4)^(2-)`

A

For disproportionation reaction.

B

For comproportionation reaction.

C

For either intermolecular redox reaction or displacement reaction

D

For either thermal combination redox reaction or thermal decomposition redox reaction.

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To solve the given reaction \( \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{SO}_3^{2-} \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-} \), we will follow these steps: ### Step 1: Determine Oxidation States First, we need to determine the oxidation states of each element in the reactants and products. 1. **Chromium in \( \text{Cr}_2\text{O}_7^{2-} \)**: - Let the oxidation state of Cr be \( x \). - The equation becomes: \( 2x + 7(-2) = -2 \) - Simplifying gives: \( 2x - 14 = -2 \) → \( 2x = 12 \) → \( x = +6 \) 2. **Sulfur in \( \text{SO}_3^{2-} \)**: - Let the oxidation state of S be \( y \). - The equation becomes: \( y + 3(-2) = -2 \) - Simplifying gives: \( y - 6 = -2 \) → \( y = +4 \) 3. **Chromium in \( \text{Cr}^{3+} \)**: - The oxidation state is \( +3 \). 4. **Sulfur in \( \text{SO}_4^{2-} \)**: - Let the oxidation state of S be \( z \). - The equation becomes: \( z + 4(-2) = -2 \) - Simplifying gives: \( z - 8 = -2 \) → \( z = +6 \) ### Step 2: Identify Changes in Oxidation States Now we can summarize the changes in oxidation states: - Chromium: \( +6 \) in \( \text{Cr}_2\text{O}_7^{2-} \) to \( +3 \) in \( \text{Cr}^{3+} \) (Reduction) - Sulfur: \( +4 \) in \( \text{SO}_3^{2-} \) to \( +6 \) in \( \text{SO}_4^{2-} \) (Oxidation) ### Step 3: Classify the Reaction Since there is a decrease in oxidation state for chromium (reduction) and an increase in oxidation state for sulfur (oxidation), this reaction is classified as a **redox reaction**. ### Step 4: Determine the Type of Redox Reaction This redox reaction involves two different species undergoing oxidation and reduction, making it an **intermolecular redox reaction**. ### Conclusion The final classification of the reaction is: - **Type of Reaction**: Intermolecular Redox Reaction
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Three different solution of oxidising agents. K_(2)Cr_(2)O_(7),I_(2), and KMnO_(4) is titrated separately with 0.19g of K_(2)S_(2)O_(3) . The molarity of each oxidising agent is 0.1 M and the reaction are: (i). Cr_(2)O_(7)^(2-)+S_(2)O_(3)^(2-)toCr^(3+)+SO_(4)^(2-) (ii). I_(2)+S_(2)O_(3)^(2-)toI^(ɵ)+S_(4)O_(6)^(2-) (iii). MnO_(4)^(ɵ)+S_(2)O_(3)^(2-)toMnO_(2)+SO_(4)^(2-) (Molecular weight of K_(2)S_(2)O_(3)=190,K_(2)Cr_(2)O_(7)=294,KMnO_(4)=158, and I_(2)=254 mol^(-1)) Which of the following statements is/are correct?

Balance the following redox reactions by ion-electron method. Cr_(2)O_(7)^(2-)+SO_(2)(g)toCr^(3+)(aq)+SO_(4)^(2-)(aq) (in acidic solution)

Three different solutions of oxidising agents KMnO_(4),K_(2)Cr_(2)O_(7) " and "I_(2) is titrated separately with 0.158 gm of Na_(2)S_(2)O_(3) . If molarity of each oxidising agent is 0.1 M and reactions are : I. MnO_(4)^(-)+S_(2)O_(3)^(2-) toMnO_(2)+SO_(4)^(2-) II CrO_(7)^(2-)+S_(2)O_(3)^(2-) to Cr^(3+)+SO_(4)^(2-) III. I_(2)+S_(2)O_(3)^(2-) to S_(4)O_(6)^(2-)+I^(-)

Calculate the potential for half-cell containing 0.10 MK_(2)Cr_(2)O_(7)(aq),0.20M Cr^(3+)(aq) and 1.0xx10^(-4)MH^(+)(aq) . The half-cell reaction is Cr_(2) O_(7)^(2-)(aq) +14H^(+) +6e^(-) to 2Cr^(3+)(aq) +7H_(2)O(l)

What is the value of x in the following equation. Cr_2O_7^(2-)+8H^(o+)+xS_2O_3^(2-)to2Cr^(3+)+3SO_4^(2-)+3S+4H_2O

Choose correct statements (s) regarding the following reactions. Cr_(2)O_(7)^(2-)(aq)+3SO_(3)^(2-)(aq)+8H^(+) rarr 2Cr^(3+)(aq)+3SO_(4)^(2-)(aq)+4H_(2)O

For the following redox reaction the coefficients of the reactants for balanced reaction are Cr_2O_7^(2-) + SO_3^(2-) + H^+ rarr Cr^(3+) + SO_4^(2-) + H_2O

Name the following radicals (a) SO_(4)^(2-) (b) HCO_(3)^(-) (c ) OH^(-) (d) Cr_(2)O_(7)^(2-)

(a) Complete the following chemical equations: (i) Cr_(2) O_(7)^(2-) (aq) + H_(2) S(g) + H^(+) (aq) to (ii) Cu^(2+)(aq) + I^(-) (aq) to (b) How would you account for the following : (i) The oxidising power of oxoanions are in the order VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt Mn O_(4)^(-) (ii) The third ionization enthalpy of manganess (Z = 25) is exceptionally high. (iii) Cr^(2+) is a stronger reducing agent than Fe^(2+) .

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