A solution contains Fe2+, Fe3+ and I– ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is +0.77 V and E° for I2/2I– = 0.536 V.
The favourable redox reaction is:
1. Fe2+ will be oxidized to Fe3+.
2. I2 will be reduced to I–.
3. There will be no redox reaction.
4. I– will be oxidized to I2.
Standard reduction potentials of the half-reactions are given below:
F2(g)F2(g) ++ 2e-→2F-(aq)2e−→2F−(aq) ;; E°E° =+2.85=+2.85 VV
Cl2(g)Cl2(g) ++ 2e-→2Cl-(aq)2e−→2Cl−(aq) ;; E°E° =+1.36=+1.36 VV
Br2(g)Br2(g) ++ 2e-→2Br-(aq)2e−→2Br−(aq) ;; E°E° =+1.06=+1.06 VV
I2(g)I2(g) ++ e-→2I-(aq)e−→2I−(aq) ;; E°E° =+0.53=+0.53 VV
The strongest oxidizing and reducing agents, respectively, are:
1. Br2Br2 and Cl-Cl−
2. Cl2Cl2 and Br-Br−
3. Cl2Cl2 and I2I2
4. F2F2 and l-l−
a | b | c | d | e | f | |
1. | 2 | 4 | 6 | 8 | 4 | 2 |
2. | 1 | 4 | 10 | 3 | 1 | 4 |
3. | 4 | 1 | 10 | 1 | 3 | 4 |
4. | 10 | 4 | 1 | 3 | 4 | 2 |
Best description of the behavior of bromine in the reaction given below is:
H2O+Br2→HOBr+HBrH2O+Br2→HOBr+HBr
1. Both oxidized and reduced
2. Oxidized only
3. Reduced only
4. Proton acceptor only
If the oxidation numbers of A, B, and C are + 2, +5, and –2 respectively, then the possible formula of the compound is:
1. | A2(BC2)2 | 2. | A3(BC4)2 |
3. | A2(BC3)2 | 4. | A3(B2C)2 |
A non-feasible reaction among the following is:
1. 22 KI+Br2→2KBr+I2KI+Br2→2KBr+I2
2. 22 KBr+I2→2KI+Br2KBr+I2→2KI+Br2
3. 22 KBr+Cl2→2KCl+Br2KBr+Cl2→2KCl+Br2
4. 2H2O+2F2→4HF+O22H2O+2F2→4HF+O2
Standard electrode potentials are:
Fe+2/FeFe+2/Fe ,E°E° == -0.44 −0.44
Fe+3/Fe+2 ,E° Fe+3/Fe+2 ,E° =0.77=0.77
Choose the correct observation when Fe+2,Fe+2, Fe+3Fe+3, and Fe (solid) are kept together:
1. Fe+3Fe+3 increases
2. Fe+3Fe+3 decreases
3. Fe+2Fe+3Fe+2Fe+3 remains unchanged
4. Fe+2Fe+2 decreases
The oxidation states(O.S.) of sulphur in the anions SO32-,SO32−, S2O42-S2O42− andand S2O62- follow the order:
1. S2O42-<SO32-<S2O62-
2. SO32-<S2O42-<S2O62-
3. S2O42-<S2O62-<SO32-
4. S2O62-<S2O42-<SO32-
From the following, identify the reaction having the top position in the EMF series (standard reduction potential) according to their electrode potential at 298 K.
1. Mg2++ 2e–→Mg(s)
2. Fe2+ + 2e–→ Fe(s)
3. Au3++ 3e–→Au(s)
4. K++ le –→K(s)
The incorrect oxidation number of the underlined atom in the following species is:
1. | Cu2O is -1 | 2. | ClO−3 is +5 |
3. | K2Cr2O7 is +6 | 4. | HAuCl4 is +3 |