Question

$H^{+}+C{r}_2{O}_7^{2-}+S{O}_3^{2-}\to C{r}^{3+}+S{O}_4^{2-}$ In a balance redox reaction, coefficient of $H^{+}$ & $H_{2}O$ will be respectively:-

• A. 3.8
• B. 3,4
• C. 4.1
• D. 8,3

Answer 1

The correct option is D 8,3

Step 1: Balancing the no of $e^{-}s$.
$\begin{array}{cc}C{r}_2{O}_7^{2-}\to 2C{r}^{3+}& S{O}_3^{2-}\to S{O}_4^{2-}\\ O.S\Rightarrow x=+6x=+3& O.Sx=+4x=+6\\ thusfor1Cr=3e^{-}& thus,fors=2e^{-}\\ 2Cr=6e^{-}& \\ thus,& thus,\\ C{r}_2{O}_7^{2-}+6e^{-}\to 2C{r}^{3+}\cdots \left(1\right)& S{O}_3^{2-}\to S{O}_4^{2-}+2e^{-}\cdots \left(2\right)\\ & Onmultiplyinge{q}^n\left(2\right)by\left(3\right)weget\\ & 3S{O}_3^{2-}\to 3S{O}_4^{2-}+6e^{-}\cdots \left(3\right)\end{array}$
On adding $e{q}^n$ (1) & (3) we get
$C{r}_2{O}_7^{2-}+H^{+}3S{O}_3^{2-}\to 2C{r}^{3+}+3S{O}_4^{2-}\cdots \left(4\right)$

Step 2: Balancing the charge-
$\begin{array}{cc}OnL.H.S& OnR.H.S\\ Totalcharge& Totalcharge\\ -7& 0\end{array}$
Thus on adding $7H^{+}$ in L.H.S in $e{q}^n$ (4) - we get
$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}+8H^{+}\to 2C{r}^{3+}+3S{O}_4^{2-}\cdots \left(5\right)$

Step 3: Balancing H & O atoms
Since in $e{q}^n$ (5) no . of Oxygen atoms.
$\begin{array}{cc}OnL.H.S& OnR.H.S\\ 16& 13\end{array}$
Thus adding $3H_{2}O$ molecules in R.H.S we get-
$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}+8H^{+}\to 2C{r}^{3+}+3S{O}_4^{2-}+3H_{2}O$
Balanced redox $e{q}^n$
Coefficient of $H^{+}\Rightarrow 8$
Coefficient of $H_{2}O\Rightarrow$