`H^+ + Cr ->H_2 + Cr^2^+`

Divide both the reaction into two half reactions.

`H^+ -> H_2`

`Cr -> Cr^2^+`

``Balance individual reactions by following oxidation state method.

`2H^+ + 2e^- -> H_2`

`Cr -> Cr^2^+ + 2e^-`

Multiply both the reaction with 2.

`(2H^+ + 2e^- -> H_2)*2` = `4H^+ + 4e^- -> 2H_2`

`(Cr -> Cr^2^+ + 2e^-) * 2` = `2Cr -> 2Cr^2^+ + 4e^-`

Add both the reactions...

`4H^+ + 2Cr -> 2H_2 + 2Cr^2^+`

This is your text book answer... But i beleive there is something missing in your question...

If you post complete question then i would be able to help you with steps.