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    The d-And-f-Block Elements
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    What happens when SO2 gas is passed through an acidified K2Cr2O7 solution?

    Sulfur dioxide gas turns acidified potassium dichromate solution from orange to green reduced chromium +4 to +3.

    K2Cr2O7 + 2H2SO4 + 3SO2 → 2Cr2(SO4)3 + K2SO4 + H2O

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    Write an ionic equation representing the oxidising property of acidified potassium permanganate solution.

    2MnO4-+16H++10I-Iodide →     2Mn2++ 5I2  + 8H2OMnO4- +8H+ +5Fe2+ →5Fe3+ +Mn2++4H2O

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    How would you convert potassium dichromate to potassium chromate?

    i)When the aqueous solution of potassium dichromate is treated with calculated amount of KOH, potassium chromate is formed.

     K2Cr7O7 + 2KOH → 2K2CrO4 + H2O

    ii) When an alkali is added to an orange red solution containing dichromate ions, a yellow solution is obtained due to the formation of chromate ions. 

     K2Cr2O7 + K2CO3 → 2 K2CrO4 + CO2

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    Give one reaction in which neutral solution of KMnO4 acts as an oxidizing agent.
    In a netural solution it acts as moderate oxidizing agent.

    2KMnO4 +H2O ---> 2KOH +2MnO2 +3O
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    Why is K2Cr2O7 generally preferred over Na2Cr2O7 in volumetric analysis although both are oxidising agents?
    K2Cr2O7 generally preferred over Na2Cr2O7 in volumetric analysis because Na2Cr2O7 is hygroscopic, hence it is difficult to prepare its standard solution for volumetric analysis, but because of no hygroscopic nature of K2Cr2O7 its standard solution can be prepared.
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