Electrochemistry

$\mathrm{The} \mathrm{electrolysis} \mathrm{of} \mathrm{conc} {\mathrm{H}}_2{\mathrm{SO}}_4 \mathrm{give} \mathrm{oleum} \left({\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8\right)$.

(A)

In reactivity series, Cu lies below Hydrogen and it is least electropositive among the given metals.

In corrosion, a metal is oxidized by loss of electrons to oxygen. Electrons released at anodic spot move through the same metal and go to another spot on the metal and reduce oxygen in the presence of ${\mathrm{H}}^{+}$ (which is believed to be available from ${\mathrm{H}}_2{\mathrm{CO}}_3$ formed due to dissolution of ${\mathrm{CO}}_2$ from air into water.

Determine the cell with the most negative ΔG°, we use:

$$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$$

For Zn|Zn²⁺(1M)||Ag⁺(1M)|Ag,

$$E_{\text{cell}}^{\circ }=0.80V-(-0.76V)=1.56V$$$$\mathrm{\Delta }{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }$$​

Since 1.56V is the highest, ΔG° is most negative for option (c)

1. Nature of electrolyte → determines type and number of ions ✅

2. Nature of solvent → affects ionization and mobility ✅

3. Concentration of electrolyte → changes number of free ions ✅

• Nature of electrode used → electrodes only allow current to pass; they don’t affect ionic conductance of the solution itself ❌

✅ Correct Answer: (B) The nature of the electrode used

The Nernst equation is given by:

$E=E°-\frac{RT}{nF}\ln Q$

Cell Reaction:

$$\begin{gathered} Mg\left(s\right)+2A{g}^{+}\left(aq\right)\rightleftharpoons 2Ag\left(s\right)+M{g}^{2+}\left(aq\right) \\ E=E_{\text{cell }}°-\frac{RT}{2F}\ln \frac{\left[M{g}^{2+}\right]}{{\left[A{g}^{+}\right]}^2} \end{gathered}$$

• At the anode (left side), oxidation occurs: $M\to M^{2+}+2e^{-}$ 𝑀→𝑀2++2𝑒− Sv6Kpe[]Sv6Kpe[]Sv6Kpe[]
• At the cathode (right side), reduction occurs: $X+2e^{-}\to X^{2-}$ 𝑋+2𝑒−→𝑋2− Sv6Kpe[]Sv6Kpe[] Sv6Kpe[[null,null,"1bc3879f-f6a6-439a-894e-2b7f19490913",null,null,0,0]]Sv6Kpe[]
Sv6Kpe[]

• Standard reduction potential for the cathode: $E^0(X/X^{2-})$ 𝐸0(𝑋/𝑋2−) Sv6Kpe[] is not given directly. The given value is $E^0(X/X^{2-})=0.34V$ 𝐸0(𝑋/𝑋2−)=0.34𝑉 Sv6Kpe[]. However, the reduction potential is typically given as $E^0(\text{oxidized}/\text{reduced})$ 𝐸0(oxidized/reduced) Sv6Kpe[]. The reaction at the cathode is $X+2e^{-}\to X^{2-}$ 𝑋+2𝑒−→𝑋2− Sv6Kpe[], which is a reduction. Let's assume the given value is the reduction potential for the cathode. So, $E_{\text{cathode}}^0=0.34V$ 𝐸0cathode=0.34𝑉 Sv6Kpe[].Sv6Kpe[]Sv6Kpe[]
• Standard reduction potential for the anode: $E^0(M^{2+}/M)=0.46V$ 𝐸0(𝑀2+/𝑀)=0.46𝑉 Sv6Kpe[]. So, $E_{\text{anode}}^0=0.46V$ 𝐸0anode=0.46𝑉 Sv6Kpe[].Sv6Kpe[] Sv6Kpe[[null,null,"01c67cc7-da83-4f94-bb74-2362ab7f0b90",null,null,0,0]]Sv6Kpe[]
Sv6Kpe[]

Oxygen gas is evolved during the electrolysis of aqueous $AgN{O}_3$ with platinum electrodes and dilute $H_{2}S{O}_4$with platinum electrodes.

The element with the most negative reduction potential is the strongest reducing agent. Among the given options, Cr will act as the reducing agent.

Concentrared Sulphuric acid can be used to obtain ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$ by the process of electrolysis.

At cathode :

$2{\mathrm{HSO}}_4^{-} \to {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8 + 2{\mathrm{e}}^{–}$

$\begin{array}{l}\mathrm{W}=\frac{\mathrm{M}}{\mathrm{nF}}\times \mathrm{It}\\ \mathrm{W}=\frac{63\times 9.6487\times 100}{2\times 96487}\\ \mathrm{W}=0.315\mathrm{g}\end{array}$

The option that is not an example of corrosion is:

B. Production of hydrogen by electrolysis of water

Explanation:

• A. Rusting of iron object is a type of corrosion where iron reacts with oxygen and moisture to form iron oxides (rust).
• C. Tarnishing of silver involves the formation of silver sulfide on the surface of silver items due to a reaction with sulfur compounds in the air.
• D. Development of green coating on copper and bronze ornaments is caused by the formation of copper carbonate or copper chloride, known as patina, which is also a form of corrosion.

Option B refers to a chemical reaction in which water is split into hydrogen and oxygen gases through electrolysis, and is not related to the process of corrosion.

We must make the reaction non-spontaneous, i.e., reverse the direction of the spontaneous redox reaction.

To do that, we need to apply an external voltage that opposes the cell’s emf — and it must be greater than $E^\circ_{\text{cell}}$Ecell∘​ to drive the reverse reaction.

$\begin{array}{l}\mathrm{W}=\frac{\mathrm{M}}{\mathrm{nF}}\times \mathrm{It}\\ \mathrm{W}=\frac{63\times 9.6487\times 100}{2\times 96487}\\ \mathrm{W}=0.315\mathrm{g}\end{array}$

A higher reduction potential means a stronger oxidising agent.

$$\begin{gathered} \mathrm{Mg}\left(\mathrm{s}\right) \to {\mathrm{Mg}}^{2+} +2{\mathrm{e}}^{-} (\mathrm{oxidation} \mathrm{half} \mathrm{cell}) \\ 2{\mathrm{Ag}}^{+} +2{\mathrm{e}}^{-} \to 2\mathrm{Ag} \left(\mathrm{s}\right) (\mathrm{reduction} \mathrm{half} \mathrm{cell}) \\ \mathrm{Net} \mathrm{cell} \mathrm{reaction} \mathrm{Mg}\left(\mathrm{s}\right) +2{\mathrm{Ag}}^{+}\to {\mathrm{Mg}}^{2+}+\mathrm{Ag}\left(\mathrm{s}\right) \\ \mathrm{Nernst} \mathrm{equation} {\mathrm{E}}_{\mathrm{cell}}= {{\mathrm{E}}^0}_{\mathrm{cell}} -\frac{\mathrm{RT}}{\mathrm{nF}}\ln \frac{\left[{\mathrm{Mg}}^{2+}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2} \end{gathered}$$

• Number of ions present

• Mobility of ions in solution

So, conductance is affected by:

1. Concentration of the electrolyte ✅ — affects ion concentration.

2. Nature of the electrolyte added ✅ — strong or weak electrolyte → degree of ionization changes.

3. Nature of the solvent used ✅ — dielectric constant and viscosity affect ionization and mobility.

• Nature of the electrode used ❌ — electrodes only act as conductors of current; they don’t influence ionic conductance of the solution itself.

$\mathrm{The} \mathrm{electrolysis} \mathrm{of} \mathrm{conc} {\mathrm{H}}_2{\mathrm{SO}}_4 \mathrm{give} \mathrm{oleum} \left({\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8\right)$.

Using Faraday's laws of electrolysis,

$W=Z\times I\times t$

Given:

$I=2A,t=30min=30\times 60sec$

$$\begin{gathered} \therefore W=\frac{M}{nF}\times I\times t \left(\because Z=\frac{M}{nF}\right) \\ W=\frac{27}{3\times 96500}\times 2\times 30\times 60 \end{gathered}$$

$W=0.336g$

A hydrogen electrode is defined by the half-cell reaction:

The Nernst equation for the hydrogen electrode is:

Since $E^\circ = 0$E∘=0 V for SHE, this simplifies to:

We are told that:

• The electrode is dipped in pure water

• We want emf = 0 (i.e., E = 0)

• Temperature = 25°C (already factored into the 0.0591 V term)

In pure water at 25°C:

Set $E = 0$E=0:

To make emf zero in pure water, hydrogen pressure must be:

But wait — the question asks:

What pressure would be required to make emf zero?

This is inverse of what we just did.

If we set pressure = 1 bar, then:

${\mathrm{\Delta G}}^{\circ }=–\mathrm{nFE}°$

= –2 × 96487 × 0.32

= –61750.4 J/mol

= –61.75 kJ/mol

The given cell reaction is:

$F{e}^{2+}\left(aq\right)+A{g}^{+}\left(aq\right)\to F{e}^{3+}\left(aq\right)+Ag\left(s\right)$

From the given data:

$$\begin{gathered} A{g}^{+}+e^{-}\to Ag , E^{\theta }=xV \\ F{e}^{3+}+3e^{-}\to Fe , E^{\theta }=zV \\ F{e}^{2+}+2e^{-}\to Fe , E^{\theta }=yV \end{gathered}$$

The given cell reaction shows $F{e}^{2+}$ is oxidized to $F{e}^{3+}$.
The corresponding reverse reaction of $F{e}^{3+}+e^{-}\to F{e}^{2+}$ is:

$F{e}^{2+}\to F{e}^{3+}+e^{-},$

$E^{\theta }=-(3z-2y)=2y-3z$

Add the Half-Reactions

$$\begin{gathered} F{e}^{2+}\to F{e}^{3+}+e^{-} , E^{\theta }=2y-3z \\ A{g}^{+}+e^{-}\to Ag , E^{\theta }=x \end{gathered}$$

Net cell reaction:

$F{e}^{2+}+A{g}^{+}\to F{e}^{3+}+Ag$

The standard cell potential is:

$$\begin{gathered} E_{cell}^{\theta }=E_{\text{cathode }}^{\theta }-E_{\text{anode }}^{\theta } \\ =\left(x\right)-(2y-3z) \\ =x+2y-3z \end{gathered}$$

In corrosion, a metal is oxidized by loss of electrons to oxygen. Electrons released at anodic spot move through the same metal and go to another spot on the metal and reduce oxygen in the presence of ${\mathrm{H}}^{+}$ (which is believed to be available from ${\mathrm{H}}_2{\mathrm{CO}}_3$ formed due to dissolution of ${\mathrm{CO}}_2$ from air into water.

In aqueous AgNO₃ (Pt electrodes) (A) and dilute H₂SO₄ (Ag electrodes) (D), water oxidizes at the anode, forming O₂. In (B) and (C), Ag dissolves or sulfate ions oxidize, preventing O₂ formation.

Correct answer: Option D (A and D).

$\mathrm{The} \mathrm{electrolysis} \mathrm{of} \mathrm{conc} {\mathrm{H}}_2{\mathrm{SO}}_4 \mathrm{give} \mathrm{oleum} \left({\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8\right)$.

Greater reduction potential means a greater tendency to reduce itself and oxidize others. Hence a stronger oxidizing agent will tend to abstract electrons easily, thus having greater reduction potential.

Statement-I is true but statement-II is false.

Greater the reduction potential , lesser will be the tendency to act as reducing agent

The cell which has most positive value of will have most negative value of.

$$\begin{gathered} {{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}^0}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}-{{\mathrm{E}}^0}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}} \\ =0.8 \mathrm{V}-(- 2.35 \mathrm{V}) =3.15 \mathrm{V} \\ ∆{\mathrm{G}}^0 =-{\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}} \end{gathered}$$

• Anode (Oxidation): $\frac{1}{2}{H}_2\left(g\right)\to H^{+}\left(aq\right)+e^{-}$ 12𝐻2(𝑔)→𝐻+(𝑎𝑞)+𝑒−
• Cathode (Reduction): $AgCl\left(s\right)+e^{-}\to Ag\left(s\right)+C{l}^{-}\left(aq\right)$ 𝐴𝑔𝐶𝑙(𝑠)+𝑒−→𝐴𝑔(𝑠)+𝐶𝑙−(𝑎𝑞)
• Overall Reaction: The sum of these two half-reactions gives the overall reaction: $\frac{1}{2}{H}_2\left(g\right)+AgCl\left(s\right)\to H^{+}\left(aq\right)+C{l}^{-}\left(aq\right)+Ag\left(s\right)$ 12𝐻2(𝑔)+𝐴𝑔𝐶𝑙(𝑠)→𝐻+(𝑎𝑞)+𝐶𝑙−(𝑎𝑞)+𝐴𝑔(𝑠) .
• Cell Notation: The cell notation $Pt\left|\right|H_2\left(g\right)\left|\right|HCl\left(soln\mathrm{.}\right)\left|\right|AgCl\left(s\right)\left|\right|Ag$ 𝑃𝑡||𝐻2(𝑔)||𝐻𝐶𝑙(𝑠𝑜𝑙𝑛.)||𝐴𝑔𝐶𝑙(𝑠)||𝐴𝑔 correctly represents this process, with the anode on the left and the cathode on the right. The $HCl$ 𝐻𝐶𝑙 solution provides both $H^{+}$ 𝐻+ and $C{l}^{-}$ 𝐶𝑙− ions for the respective half-cells.

$$\begin{gathered} {\mathrm{Ag}}^{+} + {\mathrm{e}}^{-} \to \mathrm{Ag} ; {{\mathrm{E}}_1}^{\mathrm{o}} =\mathrm{x} (\mathrm{n}=1) \\ {\mathrm{Fe}}^{2+} +2 {\mathrm{e}}^{-} \to \mathrm{Fe} ; {{\mathrm{E}}_1}^{\mathrm{o}} =\mathrm{y} ({\mathrm{n}}_1=2) \\ {\mathrm{Fe}}^{3+} +3 {\mathrm{e}}^{-} \to \mathrm{Fe} ; {{\mathrm{E}}_2}^{\mathrm{o}} =\mathrm{z} ({\mathrm{n}}_2=3) \\ {\mathrm{Fe}}^{2+} \to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-} {{\mathrm{E}}_3}^{\mathrm{o}} =? (\mathrm{n}=3) \\ \mathrm{Now} {{\mathrm{E}}_3}^{\mathrm{o}} =\frac{{\mathrm{n}}_1{{\mathrm{E}}_1}^{\mathrm{o}}+{\mathrm{n}}_2{{\mathrm{E}}_2}^{\mathrm{o}}}{\mathrm{n}} =\frac{2\mathrm{y} +3\mathrm{z}}{1}= 2\mathrm{y}+3\mathrm{z} \\ {\mathrm{Fe}}^{2+} +{\mathrm{Ag}}^{+} \to {\mathrm{Fe}}^{3+}+ \mathrm{Ag} ; \\ {\mathrm{E}}_{\mathrm{net}}^{\mathrm{o}} =\mathrm{x} + 2\mathrm{y} -3\mathrm{z} =\mathrm{x}-3\mathrm{z} +2\mathrm{y} \end{gathered}$$

The option that is not an example of corrosion is:

B. Production of hydrogen by electrolysis of water

Explanation:

• A. Rusting of iron object is a type of corrosion where iron reacts with oxygen and moisture to form iron oxides (rust).
• C. Tarnishing of silver involves the formation of silver sulfide on the surface of silver items due to a reaction with sulfur compounds in the air.
• D. Development of green coating on copper and bronze ornaments is caused by the formation of copper carbonate or copper chloride, known as patina, which is also a form of corrosion.

Option B refers to a chemical reaction in which water is split into hydrogen and oxygen gases through electrolysis, and is not related to the process of corrosion.

A. has been used in spaceship ✅
Yes, hydrogen–oxygen fuel cells were used in Apollo missions.

B. has an efficiency of 40% to produce electricity ✅
Fuel cells typically have efficiencies around 40–60%; 40% is acceptable.

C. uses aluminium as catalysts ❌
Catalyst used is platinum (Pt), not aluminium.

D. is eco-friendly ✅
Product is water, so it is environmentally friendly.

E. is actually a type of Galvanic cell only ✅
Fuel cell is a galvanic (voltaic) cell producing electrical energy from a spontaneous redox reaction.

Correct option: 🔹 D