Balance the following redox reactions in basic medium :

$B i {(O H)}_{3} + S n {(O H)}_{3}^{-} ⟶ S n {(O H)}_{6}^{2 -} + B i$

2 B i {(O H)}_{3} + 3 S n {(O H)}_{3}^{-} + 3 {O H}^{-} ⟶ 2 B i + 3 S n {(O H)}_{6}^{2 -}

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2 B i {(O H)}_{3} + 6 S n {(O H)}_{3}^{-} + 3 {O H}^{-} ⟶ 2 B i + 3 S n {(O H)}_{3}^{2 -}

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2 B i {(O H)}_{3} + 5 S n {(O H)}_{3}^{-} + 3 {O H}^{-} ⟶ 2 B i + 3 S n {(O H)}_{6}^{2 -}

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None of these

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Solution

The correct option is A $2 B i {(O H)}_{3} + 3 S n {(O H)}_{3}^{-} + 3 {O H}^{-} ⟶ 2 B i + 3 S n {(O H)}_{6}^{2 -}$
The unbalanced redox equation is as follows:
$B i {(O H)}_{3} + S n {(O H)}_{3}^{-} ⟶ S n {(O H)}_{6}^{2 -} + B i (b a s i c)$
All atoms other than H and O are balanced.
The oxidation number of Bi changes from 3 to 0. The change in the oxidation number is 3.
The oxidation number of Sn changes from 2 to 4. The change in the oxidation number is 2.
To balance the increase in the oxidation number with decrease in the oxidation number multiply $B i {(O H)}_{3}$ and $B i$ with 2 and multiply $S n {(O H)}_{3}^{-}$ and $S n {(O H)}_{6}^{2 -}$ with 3.
$2 B i {(O H)}_{3} + 3 S n {(O H)}_{3}^{-} ⟶ 3 S n {(O H)}_{6}^{2 -} + 2 B i (b a s i c)$
To balance hydroxide ions, add 3 hydroxide ions on LHS.
$2 B i {(O H)}_{3} + 3 S n {(O H)}_{3}^{-} + 3 {O H}^{-} ⟶ 2 B i + 3 S n {(O H)}_{6}^{2 -}$
This is the balanced chemical equation.

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Similar questions

C l O^{-} + C r O_{2}^{-} \to C l^{-} + C r O_{4}^{2 -}

A_{i}

| x - a_{i} | + | y | = b_{i}, i \in N,

a_{i + 1} = a_{i} + \frac{3}{2} b_{i}

b_{i + 1} = \frac{b_{i}}{2}, a_{1} = 0, b_{1} = 32,

C l O^{-} + C r O_{2}^{-} \to C l^{-} + C r O_{4}^{2 -}

K_{c}

F e_{(}^{3 +} a q) + 3 O H_{(}^{-} a q) ⇌ F e (O H)_{3 (s)}

A_{i}

| x - a_{i} | + | y | = b_{i},

a_{i + 1} = a_{i} + \frac{3}{2} b_{i}

b_{i + 1} = \frac{b_{i}}{2};

a_{1} = 0, b_{1} = 32,

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