Question

in the following chemical reaction, which element is the reducing agent? 2io₃⁻(aq) + 12h⁺(aq) + 10ag(s) + 10cl⁻(aq) → 10agcl(s) + i₂(s) answer: a i b ag c cl d h e o

Explanation:

Step1: Determine oxidation - states

In the reactants:

• For $IO_3^-$, let the oxidation - state of $I$ be $x$. Since $O$ has an oxidation - state of $- 2$ and the overall charge of $IO_3^-$ is $-1$, we have $x+3\times(-2)=-1$, so $x = + 5$.
• $Ag$ has an oxidation - state of $0$ (element in its free state).
• $Cl^-$ has an oxidation - state of $-1$.
• $H^+$ has an oxidation - state of $+1$.

In the products:

• In $AgCl$, $Ag$ has an oxidation - state of $+1$ and $Cl$ has an oxidation - state of $-1$.
• In $I_2$, $I$ has an oxidation - state of $0$.

Step2: Identify the reducing agent

A reducing agent is a substance that donates electrons and gets oxidized (increases in oxidation - state).
$Ag$ goes from an oxidation - state of $0$ in the reactant ($Ag(s)$) to $+1$ in $AgCl(s)$. So, $Ag$ is oxidized and is the reducing agent.

Answer:

B. Ag