Question

$S_4{O}_6^{2-}\left(aq\right)+Al\left(s\right)\to H_{2}S\left(aq\right)+A{l}^{3+}\left(aq\right)$

On balncing the above unblanced reaction (in acidic medium), the ratio of coefficient of $H^{+}$ to that of $H_{2}O$ is:

• A. $2:5$
• B. $10:3$
• C. $8:7$
• D. $13:5$

Answer 1

The correct option is B $10:3$

${\stackrel{+2.5}{S}}_4{O}_6^{2-}\left(aq\right)+\stackrel{0}{A}l\left(s\right)\to H_2\stackrel{-2}{S}\left(aq\right)+\stackrel{+3}{A}{l}^{3+}\left(aq\right)$

$S_4{O}_6^{2-}$ is oxidising agent.
$Al$ is reducing agent.

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

${\stackrel{+2.5}{S}}_4{O}_6^{2-}\left(aq\right)\to H_2\stackrel{-2}{S}\left(aq\right)$ oxidation
$n_f=(|-2-2.5|\times 4=18$

$\stackrel{0}{A}l\left(aq\right)\to \stackrel{+3}{A}{l}^{3+}\left(aq\right)$ reduction

$n_f=(|3-0|\times 1=3$
Ratio of $n_f$ for oxidation to reduction is $6:1$

Cross mutiply the oxidising and reducing agents with ratio of n-factors.

$S_4{O}_6^{2-}\left(aq\right)+6Al\left(s\right)\to H_{2}S\left(aq\right)+A{l}^{3+}\left(aq\right)$
Balancing elements except $O$ and $H$

$S_4{O}_6^{2-}\left(aq\right)+6Al\left(s\right)\to 4H_{2}S\left(aq\right)+6A{l}^{3+}\left(aq\right)$

Balance atoms oxygen by adding $H_{2}O$.
$S_4{O}_6^{2-}\left(aq\right)+6Al\left(s\right)\to 4H_{2}S\left(aq\right)+6A{l}^{3+}\left(aq\right)+6H_{2}O\left(l\right)$

Balance hydrogen by adding $H^{+}$
$S_4{O}_6^{2-}\left(aq\right)+6Al\left(s\right)+20H^{+}\to 4H_{2}S\left(aq\right)+6A{l}^{3+}\left(aq\right)+6H_{2}O\left(l\right)$

Balancing charge:
charge in reactant side = -2+20=+18
charge in product side = +18
so the balanced equation is:
$S_4{O}_6^{2-}\left(aq\right)+6Al\left(s\right)+20H^{+}\to 4H_{2}S\left(aq\right)+6A{l}^{3+}\left(aq\right)+6H_{2}O\left(l\right)$

The coefficient of $H^{+}$ is $20$ and coefficient of $H_{2}O$ is 6.
Desired ratio:
$20:6\Rightarrow 10:3$