Balance the redox equation occurring in acidic solution. #H^+(aq)+Cr(s)-> H_2(g)+Cr^(3+)(aq)#? Please help, thanks.
1 Answer
#6"H"^(+)(aq) + 2"Cr"(s) -> 3"H"_2(g) + 2"Cr"^(3+)(aq)#
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
#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,
