`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.