An equal volume of reducing agent is titrated separately with `1 M KMnO_(4)` in acid, neutral and alkaline medium. The volumes of `KMnO_(4)` required are `20 mL, 33.3 mL` and `100 mL` in acid, neutral and alkaline medium respectively. Find out oxidation state of `Mn` in each reaction product. Give balance equation. Find the volume of `1 M K_(2)Cr_(2)O_(7)` consumed if same volume of reductant is titrated in acid medium.

Explanation

To solve the problem step by step, we will analyze the titration of a reducing agent with `1 M KMnO4` in acidic, neutral, and alkaline media, and then determine the oxidation states of manganese in each case. Finally, we will find the volume of `1 M K2Cr2O7` consumed in acidic medium.

### Step 1: Understand the Reaction in Acidic Medium
In acidic medium, the oxidation state of manganese in `KMnO4` is +7. When it reacts with the reducing agent, it gets reduced to a lower oxidation state. Let’s denote the oxidation state of manganese in the product as `A`.

The half-reaction can be written as:
\[ \text{Mn}^{+7} + n_1 \text{e}^- \rightarrow \text{Mn}^{A} \]

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