Answer: Option B. -> Reducing action of H2S
:
B
In this reaction$H_2{O}_2$ acts as an oxidising agent and $H_{2}S$ acts as a reducing agent as sulphur gains electron and changes its oxidation state from -2 to 0. As its oxidation state increases, it reduces other reactants.

Answer: Option C. -> MnO−4/Mn2+, E0=+1.52
:
C
Reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Measured in volts (V), or millivolts (mV) - each species has its own intrinsic reduction potential. The more positive the potential, the greater the species' affinity for electrons and tendency to be reduced. Hence - among the given options - $Mn{O}_4^{-}$is strongest oxidising agent. In other words -$Mn{O}_4^{-}$ has the best affinity to get reduced and thus has the greatest ability to oxidise.

Answer: Option B. -> 2:3
:
B
The balanced equation is: $2C_2{H}_6+7O_2⟶4C{O}_2+6H_{2}O$. As you can see - the ratio of the coefficients of $C{O}_2$ and $H_{2}O$ is 4:6 or 2:3.

Answer: Option C. -> H2+Br2⟶2HBr
:
C
Observe this reaction:

In the elemental state, both $H_2$ and $B{r}_2$ have zero Oxidation numbers. In $HBr$ - the hydrogen atom has +1 oxidation state whereas the bromine atom has -1 oxidation state.

Answer: Option A. -> an oxidising agent
:
A
First - let us write down the reaction:
$S{O}_2+2H_{2}S⟶3S+2H_{2}O$
In $H_{2}S$, sulphur has an oxidation number of-2. In $S{O}_2$ Sulphur has an oxidation number of +4.
On the right-hand side, we only have elemental sulphur whoseoxidation number is 0. The sulphur in $S{O}_2$ changes its oxidation number from +4 to 0.
This means that the sulphur is gaining electrons. We know that when a reactantgains electrons,it gets reduced. Hence, $S{O}_2$ is getting reduced.
Also, the reactant that gets reduced is the oxidising agent. Hence, here, $S{O}_2$ acts as an oxidising agent.

Answer: Option B. -> Donate electrons
:
B
A substance which is capable of reducing other substances by donating electrons is called a reducing agent or reductantor reducer.

Answer: Option A. -> F2
:
A
Fluorine is the most powerful oxidising agent as reflected by a very high reduction potential of nearly 2.87.

Answer: Option A. -> C2O2−4
:
A
$C_2{O}_4^{2-}+Mn{O}_4^{-}+H^{+}⟶M{n}^{2+}+C{O}_2+H_{2}O$
In this reaction $C_2{O}_4^{2-}$ act as a reducing agent.

Answer: Option A. -> Iodine
:
A
Starch paper are used for iodine test
as: $I^{-}+oxidant⟶I_2$
$I_2+starch⟶bluecolour$

Answer: Option B. -> NaNO3
:
B
$NaN{O}_2$and $HI$ have reducing and oxidising properties respectively but$NaN{O}_3$has only oxidising property.