Electrochemistry

To determine the correct statement(s) regarding the electrochemical oxidation of hydrazine ($\mathrm{N}_2 \mathrm{H}_4$) by $\mathrm{O}_2$, let's analyze each option in detail.

Option A: $\mathrm{OH}^{-}$ ions react with $\mathrm{N}_2 \mathrm{H}_4$ at the anode to form $\mathrm{N}_2$$(\mathrm{~g})$ and water, releasing 4 electrons to the anode. This statement is accurate because in an electrochemical cell, the anode is where oxidation occurs. Hydrazine can be oxidized in the presence of $\mathrm{OH}^{-}$ ions to yield nitrogen gas and water, releasing electrons as a part of this redox reaction:

$ \mathrm{N}_2 \mathrm{H}_4 + 4 \mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4 \mathrm{H}_2 \mathrm{O} + 4e^{-} $

Option B: At the cathode, $\mathrm{N}_2 \mathrm{H}_4$ breaks to $\mathrm{N}_2$ $(\mathrm{~g})$ and nascent hydrogen released at the electrode reacts with oxygen to form water. This statement is incorrect because, usually, the cathode is where reduction occurs, not a breaking down of hydrazine. Instead, reduction reactions involve the gain of electrons. Additionally, at the cathode, it's more common for oxygen to be reduced rather than hydrazine itself breaking down.

Option C: At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$. This statement is correct. At the cathode in an alkaline medium, oxygen undergoes reduction to form hydroxide ions $\mathrm{OH}^{-}$, as depicted by the half-reaction:

$ \mathrm{O}_2 + 2H_2O + 4e^{-} \rightarrow 4 \mathrm{OH}^{-} $

Option D: Oxides of nitrogen are major by-products of the electrochemical process. This statement is incorrect because the primary product from the electrochemical reaction of hydrazine is nitrogen gas ($\mathrm{N}_2$), and not oxides of nitrogen. The formation of oxides of nitrogen would be notable only under different specific conditions not mentioned in this context.

Therefore, the correct statements regarding the electrochemical oxidation of hydrazine by $\mathrm{O}_2$ are:

Option A and Option C.

Given,

At anode : $X(s) \to \mathop {{X^{2 + }}(aq)}\limits_{(0.001\,M)} + 2{e^ - }$

At cathode : $\mathop {{Y^{2 + }}(aq)}\limits_{(0.1\,M)} + 2{e^ - } \to Y(s)$

Overall reaction $X(s) + {Y^{2 + }}(aq) \to {X^{2 + }}(aq) + Y(s)$

Nernst equation,

${E_{cell}} = E_{cell}^o - {{0.06} \over 2}\log {{[{X^{2 + }}]} \over {[{Y^{2 + }}]}}$

${E_{cell}} = E_{cell}^o - 0.03\log {{0.001} \over {0.1}}$

${E_{cell}} = E_{cell}^o + 0.06$

E_cell should be positive for a reaction to be spontaneous.

(a) $E_{C{d^{2 + }}/Cd}^o = - 0.40V,E_{N{i^{2 + }}/Ni}^o = - 0.24V$

$E_{cell}^o = E_{cathode}^o - E_{anode}^o = - 0.24 - ( - 0.40)$

$E_{cell}^o = 0.16$

$E_{cell}^{} = E_{cell}^o + 0.06 = 0.16 + 0.06 = 0.22V$

Reaction is spontaneous.

(b) $E_{C{d^{2 + }}Cd}^o = - 0.40V$ $E_{F{e^{2 + }}/Fe}^o = - 0.44V$

$E_{cell}^o = - 0.44 + 0.40 = - 0.04V$

$E_{cell}^{} = - 0.04 + 0.06 = 0.02V$

Reaction is spontaneous.

(c) $E_{N{i^{2 + }}/Ni}^o = - 0.24V$

$E_{P{b^{2 + }}/Pb}^o = - 0.13V$

$E_{cell}^o = - 0.13 + 0.24 = 0.11V$

$E_{cell}^{} = 0.11 + 0.06 = 0.17V$

Reaction is spontaneous.

(d) $E_{N{i^{2 + }}/Ni}^o = - 0.24V$

$E_{F{e^{2 + }}/Fe}^o = - 0.44V$

$E_{cell}^o = - 0.44 + 0.24 = - 0.20V$

$E_{cell}^o = - 0.20 + 0.06 = - 0.14V$

Reaction is non-spontaneous.

Therefore, the correct combinations of X and Y are (a), (b) and (c).

In a galvanic cell, also known as a voltaic cell, a salt bridge plays a crucial role in maintaining electrical neutrality within the internal circuit, which is critical for the ongoing electrochemical reaction and thus the production of electrical energy. Here, we will analyze each of the given options in the context of the function of a salt bridge in a galvanic cell:

Option A: does not participate chemically in the cell reaction.

This statement is correct. The salt bridge in a galvanic cell does not directly participate in the chemical reactions occurring at the electrodes. Its primary function is to complete the electrical circuit between the cathode and anode by allowing the transfer of ions. The salt bridge contains a salt solution (usually KNO₃, KCl, or NH₄NO₃), where the ions migrate to oppose and balance the charge buildup due to the migration of electrons through the external circuit.

Option B: stops the diffusion of ions from one electrode to another.

This option can be misleading. The salt bridge does not stop the diffusion of ions from diffusing from one electrode to another but rather provides a pathway for ions to flow back and forth, which is essential to maintain the charge balance. Hence, this statement is not the best descriptor of the salt bridge's function.

Option C: is necessary for the occurrence of the cell reaction.

This option is correct. A salt bridge is necessary for the occurrence of the cell reaction because it maintains electrical neutrality in the electrochemical cell. Without the salt bridge, the flow of electrons through the external circuit would soon cease as the solutions in the anode and cathode compartments become respectively positively and negatively charged, which would stop the electrochemical reaction.

Option D: ensures mixing of the two electrolytic solutions.

This statement is incorrect. The purpose of the salt bridge is not to mix the two electrolytic solutions; in fact, it prevents them from mixing, which could otherwise result in a direct neutralization reaction that could interfere with the proper functioning of the cell. The ions in the salt bridge only migrate enough to balance the charges in the separate solutions.

In summary, the correct answers are Option A ("does not participate chemically in the cell reaction") and Option C ("is necessary for the occurrence of the cell reaction").

The general criterion for a spontaneous redox reaction is that the reduction potential of the reducing agent (which gets oxidized) must be lower than the reduction potential of the oxidizing agent (which gets reduced). Here, the reduction of NO₃^- has an E⁰ of +0.96 V. For a metal ion to be oxidized by NO₃^-, the E⁰ of the metal ion’s reduction must be less than +0.96 V.

Let's check each given ion to see if it can be oxidized:

• V²⁺ (E⁰ = -1.19 V): This potential is significantly lower than +0.96 V, thus vanadium can be oxidized by nitrate, as it is much easier to reduce NO₃^- than to reduce V²⁺ to V.
• Fe³⁺ (E⁰ = -0.04 V): This potential is also lower than +0.96 V, so iron can be oxidized by nitrate.
• Au³⁺ (E⁰ = +1.40 V): Since the potential for Au³⁺ is higher than the potential for nitrate reduction, gold cannot be oxidized by nitrate. It means nitrate cannot provide sufficient potential to reduce Au³⁺.
• Hg²⁺ (E⁰ = +0.86 V): This potential is close but still lower than +0.96 V, therefore mercury can theoretically be oxidized by nitrate, though it is only slightly easier to reduce NO₃^- than to reduce Hg²⁺.

Given this analysis:

• Option A (V and Hg): Correct, as both V and Hg have lower reduction potentials than the nitrate ion.
• Option B (Hg and Fe): Correct, as analyzed above, both can indeed be oxidized by nitrate.
• Option C (Fe and Au): Incorrect because Au³⁺ has a higher potential and thus cannot be oxidized by nitrate.
• Option D (Fe and V): Correct, given that both V and Fe have lower potentials than nitrate reduction.

The correct answer includes Option A, Option B, and Option D.