Balance the following redox reactions:

${A u}^{3 +} (a q) + I^{-} (a q) ⟶ A u (s) + I_{2} (g)$

3 {A u}^{3 +} (a q) + 7 I^{-} (a q) ⟶ A u (s) + 3 I_{2} (g)

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2 {A u}^{3 +} (a q) + 6 I^{-} (a q) ⟶ 2 A u (s) + 3 I_{2} (g)

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4 {A u}^{3 +} (a q) + 6 I^{-} (a q) ⟶ 2 A u (s) + I_{2} (g)

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2 {A u}^{3 +} (a q) + 2 I^{-} (a q) ⟶ 2 A u (s) + 5 I_{2} (g)

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Solution

The correct option is B $2 {A u}^{3 +} (a q) + 6 I^{-} (a q) ⟶ 2 A u (s) + 3 I_{2} (g)$
The balanced reaction is as follows:
$2 {A u}^{3 +} (a q) + 6 I^{-} (a q) ⟶ 2 A u (s) + 3 I_{2} (g)$
Au is getting reduced from $+ 3$ to $0$ .
Hence, there will be a factor of $3$ before iodine in the product and hence, $6$ before iodine ion.
I is getting reduced from $- 1$ to $0$ .
According to the equation, there will be $2 \times 1$ electrons involved.
So, there will be a factor of $2$ before Au and Au ion.
Hence, option B is correct.

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