Question

W e the balanced ionic equation for the reaction of potassium dichromate (VD, K, Cr.0, with o m sulphite. Na,so, in acid medium to give chromium ion and sulphate ion. ​

Answer 1

Answer : The net ionic equation will be,

$\displaystyle{Cr_2O_7^{2-}+8H^++3SO_3^{2-}\rightarrow 3SO_4^{2-}+2Cr^{3+}+4H_2O}$

Explanation :

In the half reaction method, the number of atoms in each half reaction and number of electrons should be balanced.

The half reactions in the acidic medium are :

Reduction :    $\displaystyleCr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O ......(1)}$

Oxidation :     Oxidation : $\displaystyle{SO_3^{2-}+H_2O\rightarrow SO_4^{2-}+2H^++2e^- \ .....(2)}$

Now multiply the equation (2) by 3 and then added both equation, we get the balanced redox reaction.

Thus, the balanced redox reaction will be,

$\displaystyle{Cr_2O_7^{2-}+8H^++3SO_3^{2-}\rightarrow 3SO_4^{2-}+2Cr^{3+}+4H_2O}$

Answer 2

Answer:

$\displaystyle{Cr_2O_7^{2-}+SO_3^{2-}+8H^+\implies Cr^{3+}+SO_4^{2-}+4H_2O}$

Explanation:

Skeleton of equation we have

$\displaystyle{Cr_2O_7^{2-}+SO_3^{2-}\implies Cr^{3+}+SO_4^{2-}}$

Now , first find atom who are going in oxidation and reduction

$\displaystyle{Cr_2O_7^{2-}}\\\\\displaystyle{Cr = +6}$

$\displaystyle{SO_3^{2-}}\\\\\displaystyle{S = +4}$

$\displaystyle{Cr^{3+}}\\\\\displaystyle{Cr=+3}$

$\displaystyle{SO_4^{2-}}\\\\\displaystyle{S=+6}$

Clearly we can that S goes + 4  to + 6

Increase in O.N.  [ Oxidation ]

Cr goes + 6  to + 3

Decrease in O.N.  [ Reduction ]

Now balancing increase and decrease mutiply $\displaystyle{SO_4^{2-}}$  by 3

$\displaystyle{Cr_2O_7^{2-}+3SO_3^{2-}\implies Cr^{3+}+SO_4^{2-}}$

Balance all atoms leaving O and H

$\displaystyle{Cr_2O_7^{2-}+3SO_3^{2-}\implies 2Cr^{3+}+3SO_4^{2-}}$

Balancing O atom by adding  $\displaystyle{H_2O}$

$\displaystyle{Cr_2O_7^{2-}+SO_3^{2-}\implies Cr^{3+}+SO_4^{2-}+4H_2O}$

Balance H atom by adding H⁺ ions

$\displaystyle{Cr_2O_7^{2-}+SO_3^{2-}+8H^+\implies Cr^{3+}+SO_4^{2-}+4H_2O}$

Now it is balanced equation .