# Consider the given data: ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77;$ ${\mathrm{E}}_{{\mathrm{I}}^{-}/{\mathrm{I}}_{2}}=$ $-0.54$${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=0.80;$ ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{2+}}=$ $-0.34$${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77;$ ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{2+}}=$ $-0.34$${\mathrm{E}}_{\mathrm{Ag}/{\mathrm{Ag}}^{+}}=-0.80;$ ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77$ Using the electrode potential values given above, identify the reaction which is not feasible: 1. Fe3+(aq) and I- aq) 2. Ag+(aq) and Cu(s) 3. Fe3+(aq) and Cu(s) 4. Ag(s) and Fe3+(aq)

Subtopic:  Application of Electrode Potential |
61%
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The correct statement about the electrolysis of an aqueous solution of ${\mathrm{AgNO}}_{3}$ with Pt electrode is:

 1 Pt(s) gets oxidized at cathode whereas  Ag + ( aq )  gets reduced at anode 2 Ag+ (aq)  gets reduced at cathode and is oxidized at anode 3 Ag+ (aq) gets reduced at cathode whereas water is oxidized at anode 4 Ag (s) gets oxidized at cathode whereas  H2O  is oxidised at anode

Subtopic:  Emf & Electrode Potential |
61%
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The correct statement about electrolysis of an aqueous solution of ${\mathrm{CuCl}}_{2}$ with Pt electrode is-

 1 Cu2+  ion reduced at the cathode;  Cl-  ion oxidized at the anode 2 Cu2+  ion reduced at the anode;  Cl-  ion oxidized at the cathode 3 Cu2+  ion reduced at the cathode;  H2O  ion oxidized at the anode 4 H2O  ion reduced at the cathode;  Cl-  ion oxidized at the anode

Subtopic:  Emf & Electrode Potential |
66%
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The oxidizing agent and reducing agent in the given reaction are :

$3{\mathrm{N}}_{2}{\mathrm{H}}_{4\left(\mathrm{l}\right)}+4{\mathrm{ClO}}_{3\left(\mathrm{aq}\right)}^{-}\to$
$6{\mathrm{NO}}_{\left(\mathrm{g}\right)}+4{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}+6{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}$

1. Oxidising agent = ${\mathrm{N}}_{2}{\mathrm{H}}_{4}$; Reducing agent = ${\mathrm{ClO}}_{3}^{-}$

2. Oxidising agent = ${\mathrm{ClO}}_{3}^{-}$; Reducing agent = ${\mathrm{N}}_{2}{\mathrm{H}}_{4}$

3. Oxidising agent = ${\mathrm{N}}_{2}{\mathrm{H}}_{4}$ ; Reducing agent = ${\mathrm{N}}_{2}{\mathrm{H}}_{4}$

4. Oxidising agent = ${\mathrm{ClO}}_{3}^{-}$ ; Reducing agent = ${\mathrm{ClO}}_{3}^{-}$

Subtopic:  Oxidizing & Reducing Agents |
83%
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The oxidising agent and reducing agent in the given reaction are :

${}_{}$${\mathrm{Cl}}_{2}{\mathrm{O}}_{7\left(\mathrm{g}\right)}+4{\mathrm{H}}_{2}{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\to$
$2{\mathrm{ClO}}_{2\left(\mathrm{aq}\right)}^{-}+4{\mathrm{O}}_{2\left(\mathrm{g}\right)}+5{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}$${}_{}$

1. Oxidizing agent = H2O2; Reducing agent = ${\mathrm{Cl}}_{2}{\mathrm{O}}_{7}$

2. Oxidizing agent = ${\mathrm{Cl}}_{2}{\mathrm{O}}_{7}$; Reducing agent = ${\mathrm{H}}_{2}{\mathrm{O}}_{2}$

3. Oxidizing agent = H2O2; Reducing agent = H2O2

4. None of the above

Subtopic:  Oxidizing & Reducing Agents |
80%
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The oxidation states of the central atom in the given species are, respectively:

${\mathrm{H}}_{4}{\mathrm{P}}_{2}{\mathrm{O}}_{7}$  $\mathrm{and}$ ${\mathrm{H}}_{2}{\mathrm{S}}_{2}{\mathrm{O}}_{7}$

 1 0 and +6 2 +3 and +4 3 +4 and +2 4 +5 and +6
Subtopic:  Introduction to Redox and Oxidation Number |
90%
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KI3, H2S4O6

The oxidation numbers of iodine and sulphur in the above compounds are, respectively:

1. $$\frac{1}{3}$$ ; 4
2. 2.5 ; $$\frac{1}{3}$$
3. $$-\frac{1}{3}$$ ; 2.5
4. 2.5 ; 3

Subtopic:  Introduction to Redox and Oxidation Number |
89%
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The maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen is -

 1 9 g 2 15 g 3 12 g 4 11g
Subtopic:  Balancing of Equations |
64%
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The balanced equation for the reaction between chlorine and sulphur dioxide in water is-

 1 Cl2(s) + SO2(aq) + 2H2O(I) →2Cl-(aq) + SO42-(aq) + 4H+(aq) 2 3Cl2(s) + SO2(aq) + 2H2O(I) →Cl-(aq) + SO42-(aq) + 3H+(aq) 3 Cl2(s) + 3SO2(aq) + H2O(I) →Cl-(aq) + 2SO42-(aq) + 4H+(aq) 4 2Cl2(s) + SO2(aq) + H2O(I) →2Cl-(aq) + SO42-(aq) + 4H+(aq)

Subtopic:  Balancing of Equations |
84%
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The oxidising agent and reducing agent in the given reaction are

$5{\mathrm{P}}_{4\left(\mathrm{s}\right)}+12{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}+12{{\mathrm{HO}}^{-}}_{\left(\mathrm{aq}\right)}\to$
$8{\mathrm{PH}}_{3\left(\mathrm{g}\right)}+12{{\mathrm{HPO}}_{2}^{-}}_{\left(\mathrm{aq}\right)}$

1. Oxidising agent = ${\mathrm{P}}_{4}$; Reducing agent = ${\mathrm{P}}_{4}$

2. Oxidising agent = ${\mathrm{P}}_{4}$; Reducing agent = ${\mathrm{H}}_{2}\mathrm{O}$

3. Oxidising agent = ${\mathrm{H}}_{2}\mathrm{O}$; Reducing agent = ${\mathrm{P}}_{4}$

4. None of the above

Subtopic:  Redox Titration & Type of Redox |
69%
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