SATHEE: Related Problems with Solution

Related Problems with Solution

Problem 1 : Balance the following redox equation occurring in acidic solution:

$[C r_{2} O_{7}^{2 -} + H N O_{2} \to C r^{3 +} + N O_{3}^{-} .]$

Solution :

To balance this equation, follow these steps:

Step 1: Assign oxidation states to each element: $[C r_{2} O_{7}^{2 -} : C r^{6 +}, O^{- 2}]$ $[H N O_{2} : H^{1 +}, N^{3 +}, O^{- 2}]$ $[C r^{3 +} : C r^{3 +}]$ $[N O_{3}^{-} : N^{5 +}, O^{- 2}]$

Step 2: Write down the unbalanced equation: $[C r_{2} O_{7}^{2 -} + H N O_{2} \to C r^{3 +} + N O_{3}^{-} .]$

Step 3: Break the reaction into half-reactions for oxidation and reduction: $[O x i d a t i o n : H N O_{2} \to N O_{3}^{-}]$ $[R e d u c t i o n : C r_{2} O_{7}^{2 -} \to C r^{3 +}]$

Step 4: Balance each half-reaction: Oxidation: $[H N O_{2} \to N O_{3}^{-}]$ Add 3 electrons (e⁻) to the left side to balance the charge.

Reduction: $[C r_{2} O_{7}^{2 -} \to C r^{3 +}]$ Add 6 electrons (e⁻) to the right side to balance the charge.

Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $[6 (H N O_{2} \to N O_{3}^{-})]$ $[3 (C r_{2} O_{7}^{2 -} \to C r^{3 +})]$

Step 6: Add the balanced half-reactions to get the overall balanced equation: $[6 H N O_{2} + 3 C r_{2} O_{7}^{2 -} \to 6 N O_{3}^{-} + 3 C r^{3 +} .]$

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