Explanation

Correct answer is C

Metal cation with $(-)$ value of reduction potential $\left({\mathrm{M}}^{+3}/{\mathrm{M}}^{+2}\right)$ or with $(+)$ value of oxidation potential $\left({\mathrm{M}}^{+2}/{\mathrm{M}}^{+3}\right)$ will liberate ${\mathrm{H}}_2$

Therefore they will reduce ${\mathrm{H}}^{+}$ i. ${\mathrm{e}\mathrm{V}}^{+2}$ and ${\mathrm{Cr}}^{+2}$

Explanation

Correct answer is D

The standard electrode potential (E°) of a metal ion M⁺/M in aqueous solution is calculated under standard conditions and relates to the tendency of the M⁺ ion to gain an electron to form the neutral metal atom, M.

This is indeed a redox (reduction-oxidation) process, and the standard electrode potential is defined for the reduction half-reaction. E° values are based on the energies involved in the processes of ionization and hydration.

However, the standard electrode potential does not depend on the state (solid, liquid, gas) of the atom being ionized. Ionization of a solid metal atom (Option C) would not directly impact the electrode potential as it is not inherently part of the process of reduction at the electrode that the E° values are measuring.

The correct answer should be Option C : Ionisation of a solid metal atom.

Explanation

Correct answer is B

The provided reaction is:

$\frac{1}{2}{\mathrm{H}}_2(\mathrm{g})+\mathrm{AgCl}(\mathrm{s})\rightleftharpoons {\mathrm{H}}^{+}(\mathrm{aq})+{\mathrm{Cl}}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

This reaction involves the following half-reactions:

1. Oxidation of hydrogen gas to H+ ions: $\frac{1}{2}{\mathrm{H}}_2(\mathrm{g})\rightleftharpoons {\mathrm{H}}^{+}(\mathrm{aq})+e^{-}$


2. Reduction of AgCl to Ag: $\mathrm{AgCl}(\mathrm{s})+e^{-}\rightleftharpoons \mathrm{Ag}(\mathrm{s})+{\mathrm{Cl}}^{-}(\mathrm{aq})$

Looking at the options provided:

Option A: Doesn't involve H₂ gas, so it can't be correct.

Option B: This includes the necessary elements - H₂, AgCl, and Ag.

Option C: Doesn't involve AgCl, so it can't be correct.

Option D: Also includes the necessary elements - H₂, AgCl, and Ag.

However, looking closely, we can see that Option B represents the galvanic cell for this reaction. The reaction requires the oxidation of H₂ to H⁺, which occurs at the anode. The reaction also requires the reduction of AgCl to Ag and Cl-, which occurs at the cathode.

In Option B, the anode (on the left) is where H₂ is being oxidized to H⁺. The cathode (on the right) is where AgCl is reduced to Ag and Cl^-. The salt bridge or ion exchange component is HCl, which allows for the flow of ions to balance charge in the cell.

Therefore, the reaction occurs in the galvanic cell represented by Option B.

correct answer is 25

# Explanation

Given:

Standard reduction potential for Cu²⁺/Cu, E° = 0.34 V

Ksp of Cu(OH)₂ = 1 × 10⁻²⁰

2.303RT/F = 0.059 V

pH = 14

First, we have the solubility equilibrium for Cu(OH)₂:

$\mathrm{Cu}(\mathrm{OH}{)}_2(\mathrm{s})\rightleftharpoons {\mathrm{Cu}}^{2+}(\mathrm{aq})+2{\mathrm{OH}}^{-}(\mathrm{aq})$

The Ksp expression for this reaction is:

$\mathrm{Ksp}=\left[{\mathrm{Cu}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2$

At pH 14, the concentration of OH⁻ ions is 1 M:

$\left[{\mathrm{OH}}^{-}\right]=1\mathrm{M}$

Now we can find the concentration of Cu²⁺:

$\left[{\mathrm{Cu}}^{2+}\right]=\frac{\mathrm{Ksp}}{{\left[{\mathrm{OH}}^{-}\right]}^2}=\frac{1\times {10}^{-20}}{1^2}={10}^{-20}\mathrm{M}$

The half-cell reaction for the reduction of Cu²⁺ is:

${\mathrm{Cu}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Cu}(\mathrm{s})$

Now we can use the Nernst equation to calculate the reduction potential at pH 14:

$E=E°-\frac{0.059}{n}{\log }_{10}\frac{1}{\left[{\mathrm{Cu}}^{2+}\right]}$

Here, n = 2 (number of electrons transferred in the Cu²⁺/Cu couple).

$E=0.34-\frac{0.059}{2}{\log }_{10}\frac{1}{{10}^{-20}}$

$E=0.34-\frac{0.059}{2}\times 20$

$E=0.34-0.59$

$E=-0.25\mathrm{V}$

Thus, the reduction potential at pH 14 for the Cu²⁺/Cu couple is -0.25 V. In terms of x × 10⁻² V:

$(-)x\times {10}^{-2}\mathrm{V}=-0.25\mathrm{V}$

The value of x is 25.

correct answer is 4

# Explanation

A. ${\mathrm{E}}_{\text{cell }}$ is an intensive parameter.

This statement is correct. The standard cell potential (${\mathrm{E}}_{\text{cell}}^{\ominus }$) is an intensive parameter because it depends only on the nature of the reactants and the products, and not on the size or shape of the electrodes or the amount of material present.

B. A negative ${\mathrm{E}}^{\ominus }$ means that the redox couple is a stronger reducing agent than the ${\mathrm{H}}^{+}/{\mathrm{H}}_2$ couple.

This statement is also correct. A negative standard reduction potential (${\mathrm{E}}^{\ominus }$) for a redox couple indicates that the couple is a stronger reducing agent than the standard hydrogen electrode (${\mathrm{H}}^{+}/{\mathrm{H}}_2$), which has a standard reduction potential of 0 V.

C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

This statement is correct. The amount of electricity required for a particular oxidation or reduction reaction depends on the number of electrons involved in the reaction, which in turn depends on the stoichiometry of the electrode reaction.

D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

This statement is also correct. Faraday's laws of electrolysis state that the amount of chemical reaction at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

Therefore, all of the statements A, B, C, and D are correct.

correct answer is 2

# Explanation

In this problem, we're considering three different half-cell reactions involving lead ions.

1) The reduction of ${\mathrm{Pb}}^{2+}$ ions to lead metal :

${\mathrm{Pb}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Pb}$

The Gibbs free energy change for this process can be written as $\mathrm{\Delta }{\mathrm{G}}_1^0=-2{\mathrm{FE}}_1^0$, where ${\mathrm{E}}_1^0$ is the standard cell potential, F is Faraday's constant, and the factor of 2 is because two electrons are involved in the reaction.

2) The reduction of ${\mathrm{Pb}}^{4+}$ ions to lead metal :

${\mathrm{Pb}}^{4+}+4{\mathrm{e}}^{-}\to \mathrm{Pb}$

This reaction has a Gibbs free energy change of $\mathrm{\Delta }{\mathrm{G}}_2^0=-4{\mathrm{FE}}_2^0$.

3) The oxidation of ${\mathrm{Pb}}^{2+}$ ions to ${\mathrm{Pb}}^{4+}$ ions :

${\mathrm{Pb}}^{2+}\to {\mathrm{Pb}}^{4+}+2{\mathrm{e}}^{-}$

The Gibbs free energy change for this process is $\mathrm{\Delta }{\mathrm{G}}_3^0=-2{\mathrm{FE}}_3^0$.

We can write the third reaction as the difference between the first two reactions (i.e., reaction 2 - reaction 1). This implies that the Gibbs free energy changes for these reactions should add up accordingly :

$\mathrm{\Delta }{\mathrm{G}}_3^0=\mathrm{\Delta }{\mathrm{G}}_2^0-\mathrm{\Delta }{\mathrm{G}}_1^0$

Substituting the expressions for the Gibbs free energy changes from the half-cell reactions into this equation, we get :

$-2{\mathrm{FE}}_3^0=-4{\mathrm{FE}}_2^0-(-2{\mathrm{FE}}_1^0)=2F(2n-m)$

Solving this equation for $E_3^0$ gives :

$E_3^0=m-2n$

However, the problem statement tells us that $E_3^0$ can also be written as $m-xn$. Comparing these two expressions for $E_3^0$, we see that $x$ must be equal to 2.

So, the value of $x$ is 2.

correct answer is 1825

# Explanation

$\begin{array}{rl}& {\mathrm{FeO}}_4^{2-}\stackrel{+2.2\mathrm{V}}{⟶}{\mathrm{Fe}}^{3+}\mathrm{\Delta }{\mathrm{G}}_1=-6.6\mathrm{F}\\ \\ & {\mathrm{Fe}}^{3+}\stackrel{+0.70\mathrm{V}}{⟶}{\mathrm{Fe}}^{2+}\mathrm{\Delta }{\mathrm{G}}_2=-0.7\mathrm{F}\end{array}$

Hence for

$\begin{array}{rl}& {\mathrm{FeO}}_4^{2-}⟶{\mathrm{Fe}}^{2+}\mathrm{\Delta }\mathrm{G}=-7.3\mathrm{F}\\ \\ & =-\mathrm{nEF}\\ \\ & {\mathrm{E}}_{{\mathrm{FeO}}_4^{2-}/{\mathrm{Fe}}^{+2}}^0=\frac{-7.3\mathrm{F}}{-4\mathrm{F}}=1.825,\mathrm{n}=4\\ \\ & =1825\times {10}^{-3}\mathrm{V}\end{array}$

$n=$ electron exchange of that half cell reaction.

correct answer is 1

# Explanation

Let's go through the statements one by one:

A. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.

This statement is correct. The maximum non-expansion work that a system can perform at constant temperature and pressure is given by the change in Gibbs free energy. This is especially relevant for electrochemical reactions where this non-expansion work appears as electrical work.

B. E°cell is dependent on the pressure.

This statement is incorrect. The standard cell potential, E°cell, is not dependent on pressure. It is dependent on the nature of the reactants and products (their identities and stoichiometric ratios), as well as temperature, but not pressure. The E°cell is calculated using standard conditions, which includes a standard pressure of 1 bar or approximately 1 atm.

C. $\frac{d{E}^{\theta }\text{ cell }}{\mathrm{dT}}=\frac{{\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{S}}^{\theta }}{\mathrm{nF}}$

This statement is correct. This equation is derived from the Gibbs-Helmholtz equation, which describes the temperature dependence of the change in Gibbs free energy. In an electrochemical cell, the change in Gibbs free energy can be related to the cell potential.

D. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.

This statement is correct. In a reversible electrochemical cell, the cell potential is exactly balanced by an external, opposing potential. This ensures that the reaction proceeds at an infinitesimally slow rate, allowing the system to maintain equilibrium at all times.

So, among these four statements, only one (Statement B) is incorrect.

correct answer is 3

# Explanation

Currently no explanation available

correct answer is 3

# Explanation

${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$

$\begin{array}{rl}& \mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.059}{5}\log \frac{\left[{\mathrm{Mn}}^{2+}\right]}{\left[{\mathrm{MnO}}_4^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^8}\\ \\ & \Rightarrow 1.282=1.54-\frac{0.059}{5}\log \frac{{10}^{-3}}{{10}^{-1}\times {\left[{\mathrm{H}}^{+}\right]}^8}\\ \\ & \Rightarrow \frac{0.258\times 5}{0.059}=\log \frac{{10}^{-2}}{{\left[{\mathrm{H}}^{+}\right]}^8}\\ \\ & \Rightarrow 21.86=-2+8\mathrm{pH}\\ \\ & \therefore \mathrm{pH}=2.98\\ \\ & \simeq 3\\ \\ & \end{array}$