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This is a oxidation reduction (Redox) reaction. Here what happens is that `K_2Cr_2O_7` will reduce is oxidation state and `NaNO_2` will be oxidized. Actually the oxidation number of Cr in `K_2Cr_2O_7` will reduce and the oxidation number of N in `NaNO_2` will increase.
Oxidation number of Cr in `K_2Cr_2O_7` = +6
Oxidation number of Cr in `Cr_2(SO_4)_3` = +3
Oxidation number of N in `NaNO_2` = +3
Oxidation number of N in `NaNO_3` = +5
`Cr_2(SO_4)_3` and `NaNO_3` will be formed due to oxidation reduction reaction.
This reaction will takes place in aquas medium with acidic conditions. Therefore we add `H_2SO_4` to the media. Since the reaction takes place in a aquas medium we can add water molecules to balance the reaction.
`K_2SO_4` is formed by reaction of `SO_4^(2-)` of `H_2SO_4` with` K^+` of `K_2Cr_2O_7` . There is no redox reaction between those two.
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