Balance the following equation by oxidation number method:
`K_(2)Cr_(2)O_(7)+FeSO_(4)+H_(2)SO_(4) to Cr_(2)(SO_(4))_(3)+Fe_(2)(SO_(4))+K_(2)SO_(4)+H_(2)O`

Options

A. `Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+)rarr 6 Fe^(3+)+2Cr^(3+)+7H_(2)O`
B. `2K^(+)+Cr_(2)O_(7)^(2-)+7SO_(4)^(2-)+6Fe^(2+) rarr 3 Fe^(3+) + SO_(4)^(2-)+Cr^(3+)+K^(+)+7H_(2)O`
C. `Cr_(2)O_(7)^(2-)+2K^(+)+7H^(+)+6Fe^(2+)rarr 6 Fe^(3+) + 6 Cr^(3+) + K^(+)`
D. `Cr_(2)O_(7)^(2-)+7H^(+)+6Fe^(2+)rarr 3 Fe^(2+)+2Cr^(3+)+2K^(+) + 7H_(2)O`

Answer

A

Explanation

`K_(2)Cr_(2)O_(7)+7H_(2)SO_(4)+6FeSO_(4)rarr 3 Fe_(2)(SO_(4))_(3)+Cr_(2)(SO_(4))_(3)+7H_(2)O+K_(2)SO_(4)`
In ionic form :
`2K^(+)+Cr_(2)O_(7)^(2-)+14H^(+)+7SO_(4)^(2-)+6Fe^(2+)+6SO_(4)^(2-)rarr 6 Fe^(3+)+9SO_(4)^(2-)+2Cr^(3+)+3SO_(4)^(2-)+2K^(+)+SO_(4)^(2-)+7H_(2)O`
Eliminating the common species :
`Cr_(2)O_(7)^(2-) + 14H^(+)+6Fe^(2+)rarr 6 Fe^(3+)+2Cr^(3+)+7H_(2)O`