SCSA Chemistry Oxidation and reduction

Hydrogen fuel cells do not produce carbon dioxide ($CO_2$) during operation; they generate electricity through an electrochemical reaction rather than combustion.

While this describes the efficient energy conversion mechanism of a fuel cell, it does not explain the composition of the emissions or why they are less polluting.

This statement describes the operational requirement of a fuel cell (continuous fuel supply) but is unrelated to the environmental impact or chemical composition of its emissions.

This option incorrectly ranks S as stronger than Q. In Cell 2, Q acts as the anode (oxidized) and S as the cathode (reduced), which demonstrates that Q is a stronger reducing agent than S.

This option incorrectly identifies R as the strongest reducing agent. R produces the largest voltage when reduced by Q, indicating it has the highest reduction potential and is therefore the weakest reducing agent.

This option incorrectly ranks Q as stronger than T. In Cell 3, T acts as the anode (oxidized) and Q as the cathode (reduced), which demonstrates that T is a stronger reducing agent than Q.

While both fuels produce the same amount of greenhouse gases (1 mole of $\text{CO}_2$ per mole of fuel), they produce a different number of electrons during oxidation.

This is incorrect because they produce the same amount of greenhouse gases (1 mole of $\text{CO}_2$ per mole of fuel) and a different number of electrons.

This is incorrect because they produce the same amount of greenhouse gases, as both fuels contain exactly one carbon atom per molecule and thus yield the same amount of $\text{CO}_2$.

This is a thermal decomposition reaction where the oxidation states of Calcium (+2), Carbon (+4), and Oxygen (-2) remain unchanged.

This is a combination reaction where no elements change their oxidation states (Ca remains +2, O remains -2, H remains +1).

This is an acid-base neutralization reaction where ions exchange partners without any change in oxidation numbers.

In galvanic cells, reduction always occurs at the cathode, which is the positive electrode, while oxidation occurs at the negative anode.

Only secondary cells are rechargeable and act as electrolytic cells during the charging process; primary cells cannot be recharged.

Galvanic cells convert chemical energy into electrical energy; the conversion of electrical energy into chemical energy occurs during recharging (electrolysis), which primary cells cannot undergo.

$\text{SgO}_3$ is on the reactant side of a reduction half-reaction, meaning it accepts electrons and acts as an oxidizing agent, not a reducing agent.

While $\text{Sg}_2\text{O}_5$ can act as a reducing agent by reversing the first equation (oxidation potential of $+0.046 \text{ V}$), it does not have the most positive oxidation potential.

$\text{SgO}_2$ can act as a reducing agent by reversing the second equation, but its oxidation potential would be $-0.11 \text{ V}$, making it a relatively weak reducing agent.

Pb undergoes oxidation at the anode (its oxidation state increases from 0 to +2), making it the reducing agent, not the oxidising agent.

$\text{PbSO}_4$ is the product formed at both the anode and cathode during the discharge process, rather than a reactant acting as an oxidising agent.

Sulfuric acid ($\text{H}_2\text{SO}_4$) acts as the electrolyte that provides ions for the reaction, but it is not the species being reduced.

This is incorrect. If manganese had an oxidation state of +1, the total charge of the ion would be $+1 + 4(-2) = -7$, which does not match the actual charge of -1.

This is incorrect. While +2 is a very common oxidation state for manganese (e.g., in $\text{Mn}^{2+}$ salts), it is not the oxidation state found in the permanganate ion.

This is incorrect. Manganese is in Group 7 and has only 7 valence electrons, so its maximum possible oxidation state is +7. It cannot reach +8.

This represents the reduction half-equation during discharge, not recharge, where $Ni^{3+}$ in $NiO(OH)$ is reduced to $Ni^{2+}$ in $Ni(OH)_2$.

This represents the oxidation half-equation during discharge, where solid cadmium ($Cd$) loses electrons to form $Cd(OH)_2$.

This represents the oxidation half-equation during recharge. While it occurs during recharge, it shows a loss of electrons (oxidation), not reduction.

While Fe is oxidised, the oxidising agent is the specific species that accepts electrons. In this reaction, the copper(II) ion ($\text{Cu}^{2+}$) accepts electrons, not neutral copper (Cu).

Oxidation involves the reactant losing electrons. Here, solid Fe is oxidised, not the product $\text{Fe}^{2+}$. Additionally, the oxidising agent is the ion $\text{Cu}^{2+}$, not neutral Cu.

The species being oxidised is the reactant Fe, which loses electrons to become $\text{Fe}^{2+}$. The ion $\text{Fe}^{2+}$ is the product of oxidation, not the substance being oxidised.

Fuel cells cannot be recharged by applying an electrical current like secondary batteries; instead, they require a continuous supply of fuel and oxidant to operate.

While hydrogen fuel cells do not produce greenhouse gases directly, fuel cells that use carbon-based fuels (like methane or methanol) do produce carbon dioxide as a byproduct.

Fuel cells generate electrical energy rather than consuming it. They use chemical catalysts, not electrical energy, to overcome the activation energy barrier of the reactions.

During recharge, they operate as electrolytic cells because an external power source drives the non-spontaneous reverse reaction; they only act as galvanic cells during discharge.

This describes fuel cells. Rechargeable batteries are self-contained and have a fixed amount of reactants enclosed within the cell.

Higher temperatures typically increase the rate of unwanted side reactions, which degrades the battery faster and reduces its overall lifespan.

This combination yields a voltage of approximately $0.70\text{ V}$ ($0.34\text{ V} - (-0.36\text{ V})$). This is not the largest possible voltage because neither the most positive nor the most negative half-potentials are utilized.

Combining the reduction of copper ($+0.34\text{ V}$) and the oxidation of chromium ($-0.74\text{ V}$) yields $1.08\text{ V}$. While substantial, a larger potential difference exists using the iron half-reaction.

Both half-reactions have positive reduction potentials ($+0.34\text{ V}$ and $+0.77\text{ V}$). Combining them results in a small potential difference of only $0.43\text{ V}$.

A dense, non-porous electrode would restrict the flow of gaseous reactants to the catalytic sites, significantly decreasing the reaction rate and overall efficiency.

Electrolytes are designed to transport ions like $H^+$ or $OH^-$, not the reactants themselves. Allowing reactants to mix would cause a direct chemical reaction rather than generating an electrical current.

While the membrane must prevent gases from mixing, it cannot be completely impermeable because it must allow specific ions to pass through to complete the circuit. An impermeable membrane would stop ion flow and halt electricity generation.