If equal volumes of 1 M `KMnO_(4)` and 1 M `K_(2)Cr_(2)O_(7)` solution are used to oxidise `Fe^(2+)` in acidic medium, then `Fe^(2+)` will be oxidised

If equal volumes of 0.1 M KMnO_(4) and 0.1 M K_(2)Cr_(2)O_(7) solutions are allowed to oxidise Fe^(2+) to Fe^(3+) in acidic medium, then Fe^(2+) oxidised will be:

If equal volumes of 1 M KMnO_(4) and 1M K_(2)Cr_(2)O_(7) solutions are allowed to oxidise Fe(II) to Fe(III) in acidic medium, then Fe(II) oxidised will be

Equal volumes of 1M KMnO_4 and 1M K_2Cr_2O_7 solution are allowed to oxidise Fe^(2+) ions to Fe^(3+) ions in acidic medium. The number of moles of Fe^(2+) ions oxidised in the two cases are in the ratio:

The ratio of number of moles of KMnO_4 and K _2 Cr _2 O _7 required to oxidise 0.1mol Sn^ (+4) in acidic medium :

Ratio of moles of Fe (II) oxidised by equal volumes of equimolar KMnO_(4) and K_(2)Cr_(2)O_(7) solutions in acidic medium will be:

On oxidation with KMnO_(4) in acidic medium , SO_(2) is oxidised to

On oxidation with KMnO_(4) in acidic medium , SO_(2) is oxidised to

For a given reductant , ratio of volumes of 0.2 M KMnO_(4) and 1M K_(2)Cr_(2)O_(7) in acidic medium will be

Explain the following: K_(2)Cr_(2)O_(7) acts as a powerful oxidising agent in acidic medium.

How many gm of K_(2)Cr_(2)O_(7) is present in 1 L of its N/10 solution in acid medium ?