Redox reactions practice · Easy

Option A (electrodes) is correct because electrodes are conductors through which electricity enters or leaves an object, substance, or device.

Option A is correct because it accurately defines the equivalent weight of a reductant or oxidant as the molecular weight divided by the number of electrons lost or gained per molecule.

Option A (+6) is correct because CrO5 has a butterfly structure with two peroxide linkages, leading to +6 oxidation state for Cr.

Option A (Oxidising agent) is correct because H2SO4 acts as an oxidizing agent by accepting electrons and getting reduced.

Statement-I is incorrect, and Statement-II is correct; the oxidation state of nitrogen in N2H2 is -1, while in N3H, it's fractional (-1/3). Therefore, option D is the correct answer.

Option C (H2SO4) is correct because sulfur in H2SO4 is in its highest oxidation state (+6) and cannot be further oxidized, thus it cannot act as a reducing agent.

Option B (32) is correct because the atomic weight of the element is calculated by multiplying its equivalent weight by its valency, which is determined by the isomorphous oxychloride formula.

Option A (reduction reactions) is correct because reduction reactions involve the gain of electrons.

Option B is correct because fluorine's small size and high electronegativity due to its 7 valence electrons make it a strong oxidizing agent, readily accepting electrons.

Option A (NA) is correct because Sodium (Na) typically exhibits only one oxidation state (+1) and thus does not form multiple chlorides.

Option A ((i) and (ii)) is correct because the oxidation state of carbon in $\mathrm{C}_3 \mathrm{H}_4$ is indeed $-(4 / 3)$, and electrons are never shared in fractions.

Option A (de-electronation) is correct because oxidation involves the loss of electrons, which is also known as de-electronation.

Option D ($Ba{O_2}$) is correct because in peroxides like barium peroxide, oxygen exhibits an oxidation state of -1.

Option A (+5) is correct because the oxidation number of Cl in $ClO_3^-$ is calculated by setting the sum of oxidation numbers to the overall charge.

Option F (Fluorine) is correct because it is the most electronegative element and has a fixed oxidation state of -1, thus it cannot undergo disproportionation.

Option B (oxidation) is correct because an increase in the oxidation number of an element is defined as oxidation.

Option A (${CO}_2$) is correct because carbon is already in its highest oxidation state and cannot be further oxidized to act as a reducing agent.

Option D (F) is correct because Fluorine is the most electronegative element and always exhibits a negative oxidation state.

Option C (${P}_4+{OH}^{-} \longrightarrow {PH}_3+{H}_2 {PO}_2^{-}$) is correct because it demonstrates a disproportionation reaction where the same element undergoes both oxidation and reduction.

Option A (Both the statements are correct) is correct because the reaction involves both oxidation and reduction, and sulfur in ${{\text{S}}_2}{\text{O}}_3^{2 - }$ does form colloidal sulfur.

Option C ($\mathrm{Tl}^{3+}$ only) is correct because $\mathrm{Tl}^{3+}$ ions are less stable than $\mathrm{Tl}^{+}$ ions and readily accept electrons, thus acting as an oxidizing agent.

The correct answer is $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} \to \mathrm{N}{\mathrm{H}_3} + \mathrm{HCl}$ because there is no change in oxidation number of any element during the reaction.

Option D (-1, -1, -1) is correct because hydrogen bonded to a more electropositive metal has a -1 oxidation state.

Option B (F) is correct because fluorine is the most electronegative element and only exhibits a -1 oxidation state in its compounds.

Option A (1.4 g) is correct because 1 g of calcium produces 1.4 g of calcium oxide based on stoichiometry.

Option A ($+ 3$) is incorrect. The correct oxidation number of N in NOCl is +1. The explanation provided in the original response is incorrect. The correct approach is to consider the oxidation states of O and Cl, which are -2 and -1 respectively. Therefore, the oxidation state of N is +3. However, the given correct answer is incorrect. The correct answer should be +1.

The correct answer is C (0 to + 5) because in the reaction $B{r_2} \to BrO_3^ -$, the oxidation state of bromine changes from 0 to +5.

Option C (${\mathrm{H}_2}\mathrm{S}$) is correct because in the reaction with $H_2S$, $SO_2$ acts as an oxidizing agent, getting reduced to elemental sulfur.

Option B (Fluorine) is correct because fluorine is the most electronegative element and does not have d-orbitals available for expansion of its octet, thus it only exhibits a -1 oxidation state.

Option D ($\underline {{I_2}}$) is correct because the oxidation number of iodine changes from 0 to +5, resulting in the maximum change.