$2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)$

In the following reaction, which element is oxidized?

I^{-}

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F e

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F e^{+ 3}

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I_{2}

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Solution

The correct option is A $I^{-}$
The reaction is given below:

$2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)$

+3 -1 0 +2

$I^{⊖}$ is oxidised to $I_{2}$ .

${F e}^{3 +}$ is the oxidising agent.

${F e}^{3 +}$ is reduced to ${F e}^{2 +}$ .

$I^{-}$ is the reducing agent.

Hence, the element Zn is oxidized. Thus, option A is the correct answer.

Suggest Corrections

Similar questions

2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)

2 F e_{(a q)}^{3 +} + 2 I_{(a q)}^{-} \to 2 F e_{(a q)}^{2 +} + I_{2 (s)}

E_{c e l l}^{o} = 0.236 V

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2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (s)

Δ_{r} G_{m}^{0} = -

E_{F e^{2 +} / F e (s)}^{0} = - 0.440 V E_{F e^{3 +} / F e (s)}^{0} = - 0.036 V E_{I_{2} / 2 I^{-}}^{0} = 0.539 V F = 96500 C

2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (s)

E_{c e l l}^{\circ} = 0.236 V

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2 F e^{3 +} (a q) + S n^{2 +} (a q) \to 2 F e^{2 +} (a q) + A

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