$H_{2} S + {F e}^{3 +} ⟶ {F e}^{2 +} + S ↓ + H^{\oplus}$
1

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Solution

The correct option is A 1
The correctly balanced redox reaction using half reactions are as follows

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Q. Sum of the coefficients of all the reactants and products is

H_{2} S + {F e}^{3 +} ⟶ {F e}^{2 +} + S ↓ + H^{\oplus}

Given the standard reduction potentials:

Zn²⁺/Zn = - 0.74 V, Cl₂/Cl^- = 1.36 V, H⁺/1/2H₂ = 0 V and Fe³⁺/Fe²⁺ = 0.77 V.
The order of increasing strength as reducing agent is

Q. The redox reaction involving the reducing power of hydrogen sulphide is:

S + 2 H^{+} + 2 e^{-} \to H_{2} S, E_{S / H_{2} S}^{\circ} = + 0.14 V

Two other half equations are:

F e^{3 +} + e^{-} \to F e^{2 +} E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V

B r_{2} + 2 e^{-} \to 2 B r^{-} E_{B r_{2} / B r^{-}}^{\circ} = 1.07 V

Under standard conditions, hydrogen sulphide reacts with iron

(I I I)

ions.

If this statement is true enter 1, else enter 0.

Q. The number of redox reactions from the following equations are:
a.

H_{2} S + {F e}^{3 +} ⟶ {F e}^{2 +} + S ↓ + H^{\oplus}

I^{⊖} + I O_{3}^{⊖} + H^{\oplus} ⟶ I_{2} + H_{2} O

B i (s) + N O_{3}^{⊖} + H^{\oplus} ⟶ N O_{2} + {B i}^{3 +} + H_{2} O

I^{⊖} + O_{2} (g) + H_{2} O ⟶ I_{2} + ⊖ O H

C u (s) + {A u}^{\oplus} ⟶ A u (s) + {C u}^{2 +}

Q. Element

F e

has two oxidation states

F e^{2 +}

F e^{3 +}

, the correct order of radii of these ions will be:

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H2S+Fe3+⟶Fe2++S↓+H⊕1

The correct option is A 1The correctly balanced redox reaction using half reactions are as follows

$H_{2} S + {F e}^{3 +} ⟶ {F e}^{2 +} + S ↓ + H^{\oplus}$
1

The correct option is A 1
The correctly balanced redox reaction using half reactions are as follows