Balance the following redox equaiton by both methods.
`[Cr(OH)_(4)]^(ө)+H_(2)O_(2)rarrCrO_(4)^(2-)+H_(2)O`(basic medium)

Explanation

To balance the redox reaction \([Cr(OH)_{4}]^{\text{-}} + H_{2}O_{2} \rightarrow CrO_{4}^{2-} + H_{2}O\) in a basic medium, we will use both the ion-electron method and the oxidation number method.

### Method 1: Ion-Electron Method

**Step 1: Identify Oxidation and Reduction**
- In \([Cr(OH)_{4}]^{\text{-}}\), the oxidation state of Cr is +3.
- In \(CrO_{4}^{2-}\), the oxidation state of Cr is +6.
- Therefore, Cr is oxidized from +3 to +6 (loss of electrons).
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