Question

Balance the redox equation occurring in acidic solution. `H^+(aq)+Cr(s)-> H_2(g)+Cr^(3+)(aq)`? Please help, thanks.

Answer 1

`6"H"^(+)(aq) + 2"Cr"(s) -> 3"H"_2(g) + 2"Cr"^(3+)(aq)`

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Well, I would separate it out into half-reactions before balancing.

`"H"^(+)(aq) -> "H"_2(g)`

`"Cr"(s) -> "Cr"^(3+)(aq)`

Hydrogen atom got reduced and then formed `"H"_2(g)`. Chromium atom got oxidized into the cation. So hydrogen cation gained electrons from chromium metal.

`2"H"^(+)(aq) + 2e^(-) -> "H"_2(g)`

`"Cr"(s) -> "Cr"^(3+)(aq) + 3e^(-)`

Now that the mass and charge are balanced, recombine to cancel out the electrons.

`3(2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g))`
`ul(2("Cr"(s) -> "Cr"^(3+)(aq) + cancel(3e^(-))))`
`color(blue)(6"H"^(+)(aq) + 2"Cr"(s) -> 3"H"_2(g) + 2"Cr"^(3+)(aq))`

Why do it this way? If you did not balance charge, then you would have probably gotten:

`color(red)(2"H"^(+)(aq) + "Cr"(s) -> "Cr"^(3+)(aq) + "H"_2(g))`

However, `+2 ne +3`, so this is not balanced.