Question

balance in acidic medium

k2cr2o7 + na2so3 = cr3+ + so4(2-)

potassium dichromate + sodium sulphite = chromium + sulphate ion

Answer 1

Let us , frame the equation in acidic medium,

The reaction is ; KCr₂O₇ + Na₂SO₃ + H⁺ ---> Cr⁺³ + SO₄^2-

step 1: the skeletal equation is assigned at first as;

Cr₂O₇^2- (aq) + SO₃^2- (aq) ---> Cr⁺³ (aq)+ SO₄^2- (aq.)

Step 2: Oxidation numbers

In LHS, for Cr = +6 and for and S = +4

In RHS , For Cr = +3 for S = +6

We see that, dichromate ion has got reduced to chromium by 3 oxidation no. and sulphite ion is oxidized with in crease in +2 oxidation no.

So, to balance the oxidation no. we multiply Cr ion by 2 in RHS and sulphite ion by 3 in LHS and sulphate ion by 3 in RHS as:

Cr₂O₇^2- (aq) + 3SO₃^2- (aq) ---> 2Cr⁺³ (aq)+ 3SO₄^2- (aq.)

step 3: In acidic medium, the positive ion is not balanced so, 8 H⁺ ions are added to LHS,

Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8 H⁺ ---> 2Cr⁺³ (aq)+ 3SO₄^2- (aq.)

step 4: Now 4 water molecule is added to RHS to balance eight H⁺ ions as and the final balanced equation is given as:

Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8 H⁺ ---> 2Cr⁺³ (aq)+ 3SO₄^2- (aq.)+ 4H₂O(l)