Represent the following equation in ionic form
`K_(2)Cr_(2)O_(7)+7H_(2)SO_(4)+6FeSO_(4)=3Fe(SO_(4))_(3)+Cr_(2)(SO_(4))_(3)+7H_(2)O+K_(2)SO_(4)`

Explanation

In this equation except, `H_(2)O`, all ionic in nature. Representing these compounds in ionic forms,
`2K^(+)+Cr_(2)O_(7)^(2-)+14H^(+)+7SO_(4)^(2-)+6Fe^(2+)+6SO_(4)^(2-)to6Fe^(3+)+9SO_(4)^(2-)+2Cr^(3+)+3SO_(4)^(2-)+2K^(+)+SO_(4)^(2-)+7H_(2)O`
`2K^(+)` ions and `13SO_(4)^(2-)` ions are common on both sides, so these are cancelled. The desired ionic equation reduces, to
`Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+)=6Fe^(3+)+2Cr^(3+)+7H_(2)O`
Total charge are equal on both sides, thus, the balanced ionic equation is the same as above.