Question

Balance the following chemical equation:$\underset{\left(aq\right)}{C{r}_2{O}_7^{-2}}+\underset{\left(aq\right)}{S{O}_3^{-2}}\to \underset{\left(aq\right)}{C{r}^{+3}}+\underset{\left(aq\right)}{S{O}_4^{-2}}$ (Acidic medium)

• A. $\underset{\left(aq\right)}{C{r}_2{O}_7^{-2}}+\underset{\left(aq\right)}{3S{O}_3^{-2}}+\underset{\left(aq\right)}{H^{+}}\to \underset{\left(aq\right)}{2C{r}^{+3}}+\underset{\left(aq\right)}{3S{O}_4^{-2}}+\underset{\left(aq\right)}{4H_{2}O}$
• B. $\underset{\left(aq\right)}{C{r}_2{O}_7^{-2}}+\underset{\left(aq\right)}{S{O}_3^{-2}}+\underset{\left(aq\right)}{8H^{+}}\to \underset{\left(aq\right)}{2C{r}^{+3}}+\underset{\left(aq\right)}{3S{O}_4^{-2}}+\underset{\left(aq\right)}{4H_{2}O}$
• C. $\underset{\left(aq\right)}{C{r}_2{O}_7^{-2}}+\underset{\left(aq\right)}{3S{O}_3^{-2}}+\underset{\left(aq\right)}{8H^{+}}\to \underset{\left(aq\right)}{2C{r}^{+3}}+\underset{\left(aq\right)}{3S{O}_4^{-2}}+\underset{\left(aq\right)}{4H_{2}O}$
• D. $\underset{\left(aq\right)}{C{r}_2{O}_7^{-2}}+\underset{\left(aq\right)}{3S{O}_3^{-2}}+\underset{\left(aq\right)}{8H^{+}}\to \underset{\left(aq\right)}{C{r}^{+3}}+\underset{\left(aq\right)}{3S{O}_4^{-2}}+\underset{\left(aq\right)}{4H_{2}O}$

Answer 1

The correct option is C $\underset{\left(aq\right)}{C{r}_2{O}_7^{-2}}+\underset{\left(aq\right)}{3S{O}_3^{-2}}+\underset{\left(aq\right)}{8H^{+}}\to \underset{\left(aq\right)}{2C{r}^{+3}}+\underset{\left(aq\right)}{3S{O}_4^{-2}}+\underset{\left(aq\right)}{4H_{2}O}$

Steps for Balancing redox reactions:

1. Identify the oxidation and reduction half .
2. Find the oxidising and reducing agent.
3. Find the n-factor of oxidising and reducing agent.
4. Balance atom undergoing oxidation and reduction.
5. Cross multiply the oxidising or reducing agent with simplified n-factor values.
6. Balance atoms other than oxygen and hydrogen.
7. Balancing oxygen atoms
8. Balancing hydrogen atoms
9. Balance charge

For acidic medium:
As soon as we add x $H_{2}O$ units, we add 2x $H^{+}$ ions on the opposite side.

$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$

Following the above mentioned steps:

Balance atom undergoing oxidation and reduction.
$C{r}_2{O}_7^{-2}+S{O}_3^{-2}\to 2C{r}^{+3}+S{O}_4^{-2}$
Cross multiply the oxidising or reducing agent with simplified n-factor number,
$C{r}_2{O}_7^{-2}+3S{O}_3^{-2}\to 2C{r}^{+3}+3S{O}_4^{-2}$
Balance oxygen atoms.
$C{r}_2{O}_7^{-2}+3S{O}_3^{-2}\to 2C{r}^{+3}+3S{O}_4^{-2}++4H_{2}O$
Balance Hydrogen atoms.
$C{r}_2{O}_7^{-2}+3S{O}_3^{-2}+8H^{+}\to 2C{r}^{+3}+3S{O}_4^{-2}++4H_{2}O$
balance charge
charge in reactant side = 0
charge in product side = 0
so the balanced equation is
$C{r}_2{O}_7^{-2}+3S{O}_3^{-2}+8H^{+}\to 2C{r}^{+3}+3S{O}_4^{-2}+4H_{2}O$