Question

Consider the redox reaction giben below :
$I{O}_3^{-}\left(aq\right)+Rd\left(s\right)+H_{2}O\to Rd{O}_4^{-}\left(aq\right)+I^{-}\left(aq\right)+H^{+}$
What will be the correct coefficient of the $I{O}_3^{-}$, $H_{2}O$ and $H^{+}$ respectively.

• A. $1,5,10$
• B. $5,1,10$
• C. $7,3,6$
• D. $7,6,4$

Answer 1

The correct option is C $7,3,6$

The redox reaction is :
$I{O}_3^{-}\left(aq\right)+Rd\left(s\right)+H_{2}O\to Rd{O}_4^{-}\left(aq\right)+I^{-}\left(aq\right)+H^{+}$
$Rd$ is reducing agent.
$I{O}_3^{-}$ is oxidising agent.
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$
$\stackrel{+5}{I}{O}_3^{-}\left(aq\right)+\stackrel{0}{R}d\left(s\right)\to \stackrel{+7}{R}d{O}_4^{-}\left(aq\right)+{\stackrel{-1}{I}}^{-}\left(aq\right)$

$\stackrel{0}{R}d\left(s\right)\to \stackrel{+7}{R}d{O}_4^{-}\left(aq\right)+$ oxidation
$n_f=(|7-0|\times 1=7$
$\stackrel{+5}{I}{O}_3^{-}\left(aq\right)\to {\stackrel{-1}{I}}^{-}\left(aq\right)$ reduction
$n_f=(|-1-5|\times 1)=6$
Equalising the decrease/increase in oxidation number:
Already balanced
$I{O}_3^{-}\left(aq\right)+Rd\left(s\right)\to Rd{O}_4^{-}\left(aq\right)+I^{-}\left(aq\right)$


Cross mutiply the oxidising or reducing agent with n-factor.
$7I{O}_3^{-}\left(aq\right)+6Rd\left(s\right)\to 6Rd{O}_4^{-}\left(aq\right)+7I^{-}\left(aq\right)$

Balance atoms oxygen.
$7I{O}_3^{-}\left(aq\right)+6Rd\left(s\right)+3H_{2}O\to 6Rd{O}_4^{-}\left(aq\right)+7I^{-}\left(aq\right)$

Balance hydrogen.
$7I{O}_3^{-}\left(aq\right)+6Rd\left(s\right)+3H_{2}O\to 6Rd{O}_4^{-}\left(aq\right)+7I^{-}\left(aq\right)+6H^{+}$

balance charge
charge in reactant side = -7
charge in product side = -7
so the balanced equation is $7I{O}_3^{-}\left(aq\right)+6Rd\left(s\right)+3H_{2}O\to 6Rd{O}_4^{-}\left(aq\right)+7I^{-}\left(aq\right)+6H^{+}$