In a balance equation H₂SO₄ + xHI ⟶ H₂S + yI₂ + zH₂O, the values of x, y, z are

x= 3, y= 5, z = 2
x= 4, y= 8, z = 5
x= 8, y= 4, z = 4
x= 5, y= 3, z = 4

Solution

The value of x, y, z are 8, 4, 4 respectively hence the reaction is
H₂SO₄ + 8HI + H₂S + 4I₂ + 4H₂O

Solution

The standard electrode potentials of four elements A, B, C and D are –3.05, –1.66, –0.40 and +0.80. The highest chemical reactivity will be exhibited by :(a)A(b)B(c)C(d)D

Solution

Standard electrode potential i.e.reduction potential of A is minimum (–3.05V) i.e. its oxidation potential is maximum which implies ‘A’ is maximum reactive chemically.

The following species will not exhibit disproportionation reaction

ClO^-
\(ClO{2}^{-}\)
\(ClO{3}^{-}\)
\(ClO{4}^{-}\)

Solution

In disproportionation reaction, one element of a compound will simultaneously get reduced and oxidised. In \(ClO{4}^{-}\),
∴ oxidation number of Cl is + 7 and it cannot increase it further. So, \(ClO{4}^{-}\) will not get oxidised and so will not undergo disporportionation reaction.

Which one of the following reactions involves disproportionation?

2H₂SO₄+ Cu ⟶ CuSO₄ + 2H₂O + SO₂
As₂O₃+ 3H2S ⟶ As₂S₃ + 3H₂O
2KOH + Cl₂ ⟶ KCl +KOCl + H₂O
Ca₃P₂ + 6H₂O ⟶ 3Ca(OH)₂ + 2PH₃

Solution

\(2MnO{4}^{-}\) + 5H₂O₂ + 6H⁺ ⟶ 2Z + 5O₂ +8H₂O.
In this reaction Z is

Mn⁺²
Mn⁺⁴
MnO₂
Mn

Solution

\(2MnO{4}^{-}\) + 5H₂O₂ + 6H⁺ ⟶ 2Mn²⁺ + 5O₂ + 8H₂O

Oxygen has an oxidation state of +2 in the compound :

H₂O₂
CO₂
H₂O
F₂O

Solution

In H₂O₂ : ⇒ 2 × (+1)+ 2 × x = 0
⇒x = –1
In CO₂: ⇒ 4 + 2x = 0 ⇒ x = –2
In H₂O :⇒2 × (+1) + x = 0 ⇒x = –2
In F₂O :⇒2 × (–1) + x = 0 ⇒ x = +2

C₂H₆ (g) + nO₂(g) ⟶ CO₂(g) + H₂O(l)In this equation, the ratio of the coefficients of CO₂ and H₂O is

1 : 1
2 : 3
3 : 2
1 : 3

Solution

The balanced equation is 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O.
Ratio of the coefficients of CO₂ and H₂O is 4 : 6 or 2 : 3

For the reaction : NH₃ + OCl^– ⟶ N₂H₄ + Cl^– in basic medium, the coefficients of NH₃, OCl^– and N₂H₄ for the balanced equation are respectively

2, 2, 2
2, 2, 1
2, 1, 1
4, 4, 2

Solution

The balanced equation
NH₃ + OCl^- ⟶ N₂H₄ + Cl^- + H₂

A metal ion M³⁺ loses 3 electrons, its oxidation number will be

+3
+6
zero
–3

Solution

M³⁺ on losing 3 elections will become M⁺⁶ and O.N. = + 6

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