Why is this redox reaction not balanced? Please balance it. $"H"_2"SO"_4(aq) + "K"_2"Cr"_2"O"_7(aq) + "SO"_2(g) -> "Cr"_2"SO"_4(aq) + "H"_2"O"(l) + "K"_2"SO"(aq)$

Md Ariful Islam

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Md Ariful Islam

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$SO_2(g) +2H_2O(l) rarr SO_4^(2-)+4H^(+) + 2e^(-)$ $(i)$

And dichromate, $stackrel(+VI)"Cr"_2O_7^(2-)$ is reduced to $Cr^(3+)$.

$Cr_2O_7^(2-) +14H^(+) + 6e^(-) rarr 2Cr^(3+)+7H_2O(l)$ $(ii)$

And these are both balanced with respect to mass and charge, as they must be...........and so we takes $3xx(i) + (ii)$ to eliminate the electrons.......

$3SO_2(g) +cancel(6H_2O(l)) +Cr_2O_7^(2-) +2cancel(14)H^(+) + cancel(6e^(-))rarr 3SO_4^(2-)+cancel(12H^(+)) + cancel(6e^(-))+2Cr^(3+)+cancel7H_2O(l)$

And cancel out the common reagents to give........

$Cr_2O_7^(2-)+3SO_2(g) +2H^(+)rarr 2Cr^(3+)+3SO_4^(2-)+H_2O(l)$

Which I think is balanced with respect to mass and charge, as indeed it must be if it is to reflect chemical reality.

All these take a bit a practice. What you would observe in this reaction (and I hope that it will be demonstrated in a lab) is the orange-red colour of dichromate dissipate to give green $Cr^(3+)$ ion.

See for other examples.

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