$2 {F e}^{3 +} (a q) + 2 I^{⊖} (a q) ⟶ I_{2} (a q) + 2 {F e}^{2 +} (a q)$
1

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Solution

The correct option is A 1
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 $I_{2}$ .
Therefore, ${F e}^{3 +}$ is the oxidising agent (oxidiser).
${F e}^{3 +}$ is reduced to ${F e}^{2 +}$ .
Hence, $I^{-}$ is the reducing agent (reducer).

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