Question

decide whether each chemical reaction in the table below is an oxidation - reduction (
edox\) reaction. if the reaction is a redox reaction, write down the formula of the reducing agent and the formula of the oxidizing agent.
$hso_{4}^{-}(aq)+oh^{-}(aq)
ightarrow so_{4}^{2 - }(aq)+h_{2}o(l)$
$cuso_{4}(aq)+zn(s)
ightarrow znso_{4}(aq)+cu(s)$
$2h_{2}s(g)+o_{2}(g)
ightarrow 2s(s)+2h_{2}o(l)$

Explanation:

Step1: Determine oxidation - states

For the reaction $\mathrm{HSO}_{4}^{-}(aq)+\mathrm{OH}^{-}(aq)
ightarrow\mathrm{SO}_{4}^{2 -}(aq)+\mathrm{H}_{2}\mathrm{O}(l)$:
In $\mathrm{HSO}_{4}^{-}$, hydrogen has an oxidation - state of + 1, oxygen has an oxidation - state of - 2. Let the oxidation - state of sulfur be $x$. Then $+1 + x+4\times(-2)=-1$, so $x = + 6$. In $\mathrm{SO}_{4}^{2 -}$, with oxygen at - 2, if the oxidation - state of sulfur is $y$, then $y + 4\times(-2)=-2$, so $y = + 6$. Oxidation states of all elements remain the same. So it is not a redox reaction.
For the reaction $\mathrm{CuSO}_{4}(aq)+\mathrm{Zn}(s)
ightarrow\mathrm{ZnSO}_{4}(aq)+\mathrm{Cu}(s)$:
In $\mathrm{CuSO}_{4}$, copper has an oxidation - state of + 2, in $\mathrm{Cu}(s)$ it is 0. Zinc has an oxidation - state of 0 in $\mathrm{Zn}(s)$ and + 2 in $\mathrm{ZnSO}_{4}$. Zinc is oxidized (loses electrons, oxidation state increases from 0 to + 2) and copper is reduced (gains electrons, oxidation state decreases from + 2 to 0).
For the reaction $2\mathrm{H}_{2}\mathrm{S}(g)+\mathrm{O}_{2}(g)
ightarrow2\mathrm{S}(s)+2\mathrm{H}_{2}\mathrm{O}(l)$:
In $\mathrm{H}_{2}\mathrm{S}$, sulfur has an oxidation - state of - 2, in $\mathrm{S}(s)$ it is 0. In $\mathrm{O}_{2}$, oxygen has an oxidation - state of 0, in $\mathrm{H}_{2}\mathrm{O}$ it is - 2. Sulfur is oxidized (oxidation state increases from - 2 to 0) and oxygen is reduced (oxidation state decreases from 0 to - 2).

Step2: Identify reducing and oxidizing agents

In $\mathrm{CuSO}_{4}(aq)+\mathrm{Zn}(s)
ightarrow\mathrm{ZnSO}_{4}(aq)+\mathrm{Cu}(s)$, the reducing agent is the species that gets oxidized. Since $\mathrm{Zn}$ is oxidized, $\mathrm{Zn}$ is the reducing agent. The oxidizing agent is the species that gets reduced, so $\mathrm{CuSO}_{4}$ is the oxidizing agent.
In $2\mathrm{H}_{2}\mathrm{S}(g)+\mathrm{O}_{2}(g)
ightarrow2\mathrm{S}(s)+2\mathrm{H}_{2}\mathrm{O}(l)$, the reducing agent is $\mathrm{H}_{2}\mathrm{S}$ (because sulfur in $\mathrm{H}_{2}\mathrm{S}$ is oxidized) and the oxidizing agent is $\mathrm{O}_{2}$ (because oxygen in $\mathrm{O}_{2}$ is reduced).

Answer:

For $\mathrm{HSO}_{4}^{-}(aq)+\mathrm{OH}^{-}(aq)
ightarrow\mathrm{SO}_{4}^{2 -}(aq)+\mathrm{H}_{2}\mathrm{O}(l)$:
redox reaction? no
reducing agent: N/A
oxidizing agent: N/A

For $\mathrm{CuSO}_{4}(aq)+\mathrm{Zn}(s)
ightarrow\mathrm{ZnSO}_{4}(aq)+\mathrm{Cu}(s)$:
redox reaction? yes
reducing agent: $\mathrm{Zn}$
oxidizing agent: $\mathrm{CuSO}_{4}$

For $2\mathrm{H}_{2}\mathrm{S}(g)+\mathrm{O}_{2}(g)
ightarrow2\mathrm{S}(s)+2\mathrm{H}_{2}\mathrm{O}(l)$:
redox reaction? yes
reducing agent: $\mathrm{H}_{2}\mathrm{S}$
oxidizing agent: $\mathrm{O}_{2}$