Electrochemistry
The reaction;
$\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}+\mathrm{AgCl}_{(\mathrm{s})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}+\mathrm{Ag}_{(\mathrm{s})}$
occurs in which of the following galvanic cell :
Given below are two statements :
Statement (I) : Fusion of $\mathrm{MnO}_2$ with $\mathrm{KOH}$ and an oxidising agent gives dark green $\mathrm{K}_2 \mathrm{MnO}_4$.
Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.
In the light of the above statements, choose the correct answer from the options given below :
How can an electrochemical cell be converted into an electrolytic cell ?
A conductivity cell with two electrodes (dark side) are half filled with infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will -
The quantity of silver deposited when one coulomb charge is passed through $\mathrm{AgNO}_3$ solution :
For the electro chemical cell
$\mathrm{M}\left|\mathrm{M}^{2+}\right||\mathrm{X}| \mathrm{X}^{2-}$
If $\mathrm{E}_{\left(\mathrm{M}^{2+} / \mathrm{M}\right)}^0=0.46 \mathrm{~V}$ and $\mathrm{E}_{\left(\mathrm{x} / \mathrm{x}^{2-}\right)}^0=0.34 \mathrm{~V}$.
Which of the following is correct?
Molar ionic conductivities of divalent cation and anion are $57 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}$ and $73 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:
The reaction at cathode in the cells commonly used in clocks involves.
Fuel cell, using hydrogen and oxygen as fuels,
A. has been used in spaceship
B. has as efficiency of $40 \%$ to produce electricity
C. uses aluminum as catalysts
D. is eco-friendly
E. is actually a type of Galvanic cell only
Choose the correct answer from the options given below:
For a strong electrolyte, a plot of molar conductivity against (concentration) ${ }^{1 / 2}$ is a straight line, with a negative slope, the correct unit for the slope is
One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?
What pressure (bar) of $\mathrm{H}_2$ would be required to make emf of hydrogen electrode zero in pure water at $25^{\circ} \mathrm{C}$ ?
Identify the factor from the following that does not affect electrolytic conductance of a solution.
Alkaline oxidative fusion of $\mathrm{MnO}_2$ gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are
Reduction potential of ions are given below:
$\begin{array}{ccc} \mathrm{ClO}_4^{-} & \mathrm{IO}_4^{-} & \mathrm{BrO}_4^{-} \\ \mathrm{E}^{\circ}=1.19 \mathrm{~V} & \mathrm{E}^{\circ}=1.65 \mathrm{~V} & \mathrm{E}^{\circ}=1.74 \mathrm{~V} \end{array}$
The correct order of their oxidising power is :
Which of the following statements is not correct about rusting of iron?
(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.
(B) The number of ions per unit volume that carry current in a solution increases on dilution.
(C) Molar conductivity increases with decrease in concentration
(D) The variation in molar conductivity is different for strong and weak electrolytes
(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.
Explanation:
(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.
This statement is true. Conductivity is defined as the ability of a solution to conduct electricity, and it depends on the concentration of ions in the solution. As the concentration of ions decreases, the conductivity of the solution also decreases, regardless of whether the electrolyte is strong or weak.
(B) The number of ions per unit volume that carry current in a solution increases on dilution.
This statement is false. The number of ions per unit volume in a solution decreases on dilution, because the total number of ions in the solution remains the same, but the volume increases.
(C) Molar conductivity increases with decrease in concentration.
This statement is generally true. Molar conductivity is a measure of the ability of an electrolyte to conduct electricity, and it depends on both the concentration of the electrolyte and the mobility of the ions. As the concentration of the electrolyte decreases, the molar conductivity increases because the ions become more separated from each other, and the electrostatic interactions between them become weaker, allowing them to move more freely in the solution.
(D) The variation in molar conductivity is different for strong and weak electrolytes.
This statement is true. The variation in molar conductivity with concentration is different for strong and weak electrolytes. Strong electrolytes dissociate completely in solution, so their molar conductivity increases rapidly with dilution, while weak electrolytes dissociate only partially, so their molar conductivity increases more slowly with dilution.
(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.
This statement is false. For weak electrolytes, the change in molar conductivity with dilution is due to an increase in the degree of dissociation, not a decrease.
Therefore, there are $\boxed{3}$ correct statements in the given options, namely (A), (C), (D).
At $298 \mathrm{~K}$, the standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ electrode is $0.34 \mathrm{~V}$.
Given : $\mathrm{K}_{\mathrm{sp}} \mathrm{Cu}(\mathrm{OH})_{2}=1 \times 10^{-20}$
Take $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$
The reduction potential at $\mathrm{pH}=14$ for the above couple is $(-) x \times 10^{-2} \mathrm{~V}$. The value of $x$ is ___________
Explanation:
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:
$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$
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.
A metal surface of $100 \mathrm{~cm}^{2}$ area has to be coated with nickel layer of thickness $0.001 \mathrm{~mm}$. A current of $2 \mathrm{~A}$ was passed through a solution of $\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}$ for '$\mathrm{x}$' seconds to coat the desired layer. The value of $\mathrm{x}$ is __________. (Nearest integer) ( $\rho_{\mathrm{Ni}}$ (density of Nickel) is $10 \mathrm{~g} \mathrm{~mL}$, Molar mass of Nickel is $60 \mathrm{~g} \mathrm{~mol}^{-1}$ $\left.\mathrm{F}=96500 ~\mathrm{C} ~\mathrm{mol}^{-1}\right)$
Explanation:
The formula for Faraday's law of electrolysis is:
$W = z \times i \times t$
where W is the amount of substance deposited (in grams), z is the electrochemical equivalent (grams per coulomb), i is the current (in amperes), and t is the time (in seconds).
By relating the density and volume of the nickel layer to the electric charge passed through the electrolyte, we can calculate the time needed for the deposition:
$10 \times 100 \times 0.0001 = \frac{\left(\frac{\text { atomic wt. }}{\text { v.f }}\right) \times 2 \times x}{96500}$
where v.f is the valence factor for the reaction (in this case, 2).
Solving for x, we get:
$x = 161 \, \mathrm{sec}$
So, the value of x is 161 seconds.
The number of correct statements from the following is __________
A. $\mathrm{E_{\text {cell }}}$ is an intensive parameter
B. A negative $\mathrm{E}^{\ominus}$ means that the redox couple is a stronger reducing agent than the $\mathrm{H}^{+} / \mathrm{H}_{2}$ couple.
C. The amount of electricity required for oxidation or reduction 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.
Explanation:
A. $\mathrm{E_{\text {cell }}}$ is an intensive parameter.
This statement is correct. The standard cell potential ($\mathrm{E}^{\ominus}_{\text{cell}}$) 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.
In an electrochemical reaction of lead, at standard temperature, if $\mathrm{E}^{0}\left(\mathrm{~Pb}^{2+} / \mathrm{Pb}\right)=\mathrm{m}$ Volt and $\mathrm{E}^{0}\left(\mathrm{~Pb}^{4+} / \mathrm{Pb}\right)=\mathrm{n}$ Volt, then the value of $\mathrm{E}^{0}\left(\mathrm{~Pb}^{2+} / \mathrm{Pb}^{4+}\right)$ is given by $\mathrm{m-x n}$. The value of $\mathrm{x}$ is ___________. (Nearest integer)
Explanation:
1) The reduction of $\mathrm{Pb}^{2+}$ ions to lead metal :
$\mathrm{Pb}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$
The Gibbs free energy change for this process can be written as $\Delta \mathrm{G}_1^0=-2 \mathrm{FE}_1^0$, 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 :
$\mathrm{Pb}^{4+} + 4 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$
This reaction has a Gibbs free energy change of $\Delta \mathrm{G}_2^0=-4 \mathrm{FE}_2^0$.
3) The oxidation of $\mathrm{Pb}^{2+}$ ions to $\mathrm{Pb}^{4+}$ ions :
$\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+} + 2 \mathrm{e}^{-}$
The Gibbs free energy change for this process is $\Delta \mathrm{G}_3^0=-2 \mathrm{FE}_3^0$.
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 :
$\Delta \mathrm{G}_3^0=\Delta \mathrm{G}_2^0 - \Delta \mathrm{G}_1^0$
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.
The specific conductance of $0.0025 ~\mathrm{M}$ acetic acid is $5 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$ at a certain temperature. The dissociation constant of acetic acid is __________ $\times ~10^{-7}$ (Nearest integer)
Consider limiting molar conductivity of $\mathrm{CH}_{3} \mathrm{COOH}$ as $400 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
Explanation:
Given that the specific conductance, $k$, is $5 \times 10^{-5} ~S~cm^{-1}$ and the concentration, $C$, is $0.0025~M$, we can find the molar conductivity, $\lambda_m$, as follows:
$\lambda_m = \frac{k}{C} \times 1000 = \frac{5 \times 10^{-5} \times 10^3}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 ~S~cm^2~mol^{-1}$
Next, we find the degree of dissociation, $\alpha$, by dividing $\lambda_m$ by the limiting molar conductivity, $\lambda_m^o$:
$\alpha = \frac{20}{400} = \frac{1}{20}$
Finally, we use the formula for the dissociation constant of a weak acid, $K_a$:
$K_a = \frac{C \alpha^{2}}{1-\alpha} = \frac{0.0025 \times \frac{1}{20} \times \frac{1}{20}}{\frac{19}{20}} = \frac{0.0025}{19 \times 20} = 6.6 \times 10^{-6} = 66 \times 10^{-7}$
So, the correct answer is 66.
$\mathrm{FeO_4^{2 - }\buildrel { + 2.2V} \over \longrightarrow F{e^{3 + }}\buildrel { + 0.70V} \over \longrightarrow F{e^{2 + }}\buildrel { - 0.45V} \over \longrightarrow F{e^0}}$
$E_{FeO_4^{2 - }/F{e^{2 + }}}^\theta $ is $x \times {10^{ - 3}}$ V. The value of $x$ is _________
Explanation:
Hence for
$ \begin{aligned} & \mathrm{FeO}_4^{2-} \longrightarrow \mathrm{Fe}^{2+} \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{aligned} $
$n=$ electron exchange of that half cell reaction.
The number of incorrect statements from the following is ___________.
A. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.
B. $\mathrm{E_{cell}^{\circ}}$ cell is dependent on the pressure.
C. $\frac{d E^{\theta} \text { cell }}{\mathrm{dT}}=\frac{\Delta_{\mathrm{r}} \mathrm{S}^{\theta}}{\mathrm{nF}}$
D. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.
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. $\frac{d E^{\theta} \text { cell }}{\mathrm{dT}}=\frac{\Delta_{\mathrm{r}} \mathrm{S}^{\theta}}{\mathrm{nF}}$
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.
The standard reduction potentials at $298 \mathrm{~K}$ for the following half cells are given below:
$\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O} \quad \mathrm{E}^{\theta}=0.97 \mathrm{~V}$
$\mathrm{V}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{V} \quad\quad\quad \mathrm{E}^{\theta}=-1.19 \mathrm{~V}$
$\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad\quad\quad \mathrm{E}^{\theta}=-0.04 \mathrm{~V}$
$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \quad\quad\quad \mathrm{E}^{\theta}=0.80 \mathrm{~V}$
$\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}(\mathrm{s}) \quad\quad\quad \mathrm{E}^{\theta}=1.40 \mathrm{~V}$
The number of metal(s) which will be oxidized by $\mathrm{NO}_{3}^{-}$ in aqueous solution is __________.
Explanation:
To determine the metals that can be oxidized by $NO_3^-$, we need to find the metals that have a standard reduction potential (SRP) lower than 0.97 V (since $NO_3^-$ has an SRP of 0.97 V).
From the given data, we can find the metals that satisfy this condition:
- $V^{2+} (\text{aq}) + 2e^- \rightarrow V$, $E^\theta = -1.19 \, \text{V}$ (less than 0.97 V, hence can be oxidized)
- $Fe^{3+} (\text{aq}) + 3e^- \rightarrow Fe$, $E^\theta = -0.04 \, \text{V}$ (less than 0.97 V, hence can be oxidized)
- $Ag^+ (\text{aq}) + e^- \rightarrow Ag (\text{s})$, $E^\theta = 0.80 \, \text{V}$ (less than 0.97 V, hence can be oxidized)
- $Au^{3+} (\text{aq}) + 3e^- \rightarrow Au (\text{s})$, $E^\theta = 1.40 \, \text{V}$ (greater than 0.97 V, hence cannot be oxidized)
Therefore, $V$, $Fe$, and $Ag$ are the metals that can be oxidized by $NO_3^-$.
The total number of such metals is 3.
$1 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3}$ is added to $1 \mathrm{~L}$ of saturated solution of $\mathrm{AgBr}$. The conductivity of this solution at $298 \mathrm{~K}$ is _____________ $\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1}$.
[Given : $\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13}$ at $298 \mathrm{~K}$
$ \begin{aligned} & \lambda_{\mathrm{Ag}^{+}}^{0}=6 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{Br}^{-}}^{0}=8 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \left.\lambda_{\mathrm{NO}_{3}^{-}}^{0}=7 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\right] \end{aligned} $
Explanation:
At what pH, given half cell $\mathrm{MnO_{4}^{-}(0.1~M)~|~Mn^{2+}(0.001~M)}$ will have electrode potential of 1.282 V? ___________ (Nearest Integer)
Given $\mathrm{E_{MnO_4^ - |M{n^{2 + }}}^o}=1.54~\mathrm{V},\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059\mathrm{V}$
Explanation:
$ \begin{aligned} & \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{aligned} $
Its molar conductivity is _________ $\times 10^{4}~ \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. (Nearest integer)
Explanation:
The logarithm of equilibrium constant for the reaction $\mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_{4}^{2-}$ is ___________ (Nearest integer)
Given : $\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V}$
$ \mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V} $
$ \begin{aligned} & \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned} $
Explanation:
$ \begin{aligned} & -\mathrm{nFE}_{\text {cell }}^{\mathrm{o}}=-\mathrm{RT} \times 2.303\left(\log _{10} \mathrm{~K}\right) \\\\ & \frac{\mathrm{E}_{\text {Cell }}^{\mathrm{o}}}{0.06} \times \mathrm{n}=\log \mathrm{K} .......(1) \\\\ & \mathrm{Pd}^{+2} \text { (aq.) }+ \mathrm{2e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}), \mathrm{E}_{\text {cat, }{ reduction}}^{\mathrm{o}}=0.83 \\\\ & \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-} \text {(aq.) } \rightleftharpoons \mathrm{PdCl}_4^{2-}, \text { (aq) }+2 \mathrm{e}^{-}, \mathrm{E}_{\text {Anode, Oxidation }}^0=0.65 \end{aligned} $
Net Reaction $\rightarrow \mathrm{Pd}^{2+}$ (aq.) $+4 \mathrm{Cl}^{-}$(aq.) $\rightleftharpoons \mathrm{PdCl}_4^{2-}$ (aq.)
$ \begin{aligned} & \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cat,red }}^{\mathrm{o}}-\mathrm{E}_{\text {Anded, oxidid }}^0 \\\\ & \mathrm{E}_{\text {cell }}^{\mathrm{o}}=0.83-0.65 \\\\ & \mathrm{E}_{\text {cell }}^0=0.18 .........(2) \end{aligned} $
Also $\mathrm{n}=2$ .......(3)
Using equation (1), (2) and (3)
$\log K=6$
$\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}$ is _______ $\times 10^{-2} \mathrm{~V}$ (Nearest integer)
Given: $\mathrm{E}^{0} _ {\mathrm{X}^{2+} \mid \mathrm{X}}=-2.36 \mathrm{~V}$
$\mathrm{E}_{\mathrm{Y}^{2+} \mid \mathrm{Y}}^{0}=+0.36 \mathrm{~V}$
$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$
Explanation:
Consider the cell
$\mathrm{Pt}_{(\mathrm{s})}\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{~atm})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M})|| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \operatorname{Pt}(\mathrm{s})$
When the potential of the cell is $0.712 \mathrm{~V}$ at $298 \mathrm{~K}$, the ratio $\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right]$ is _____________. (Nearest integer)
Given : $\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771$
$ \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V} $
Explanation:
Anode ${H_2} \to 2{H^ + } + 2{e^ - }$
Cathode $(F{e^{3 + }} + {e^ - } \to F{e^{2 + }}) \times 2$
${H_2} + 2F{e^{3 + }} \to 2{H^ + } + 2F{e^{2 + }}$
${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$
$0.712 = 0.771 - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$
$ - 0.059 = - 0.059\log {{F{e^{2 + }}} \over {F{e^{3 + }}}}$
${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = 10$
The equilibrium constant for the reaction
$\mathrm{Zn(s)+Sn^{2+}(aq)}$ $\rightleftharpoons$ $\mathrm{Zn^{2+}(aq)+Sn(s)}$ is $1\times10^{20}$ at 298 K. The magnitude of standard electrode potential of $\mathrm{Sn/Sn^{2+}}$ if $\mathrm{E_{Z{n^{2 + }}/Zn}^\Theta = - 0.76~V}$ is __________ $\times 10^{-2}$ V. (Nearest integer)
Given : $\mathrm{\frac{2.303RT}{F}=0.059~V}$
Explanation:
$E_{cell}^o = {{2.303\,RT} \over {2F}}\log k$
$E_{cell}^o = {{0.059} \over 2}\log ({10^{20}})$
$E_{Z{n^{2 + }}/Zn}^o + 0.76 = 0.59$
$E_{Z{n^{2 + }}/Zn}^o = 0.59 - 0.76$
$E_{Zn/Z{n^{2 + }}}^o = 0.17\,V$
Following figure shows dependence of molar conductance of two electrolytes on concentration. $\Lambda \mathop m\limits^o $ is the limiting molar conductivity.
The number of $\mathrm{\underline {incorrect} }$ statement(s) from the following is ___________
(A) $\Lambda \mathop m\limits^o $ for electrolyte A is obtained by extrapolation
(B) For electrolyte B, $\Lambda \mathop m\limits $ vs $\sqrt c$ graph is a straight line with intercept equal to $\Lambda \mathop m\limits^o $
(C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B.
(D) $\Lambda \mathop m\limits^o $ for any electrolyte A and B can be calculated using $\lambda^\circ$ for individual ions
Explanation:
(C) At infinite dilution, value of degree of dissociation approaches one.
$\therefore \mathrm{A}$ and $\mathrm{C}$ are incorrect
$Pt(s)|{H_2}(g)(1\,bar)|{H^ + }(aq)(1\,M)||{M^{3 + }}(aq),{M^ + }(aq)|Pt(s)$
The $\mathrm{E_{cell}}$ for the given cell is 0.1115 V at 298 K when ${{\left[ {{M^ + }(aq)} \right]} \over {\left[ {{M^{3 + }}(aq)} \right]}} = {10^a}$
The value of $a$ is ____________
Given : $\mathrm{E_{{M^{3 + }}/{M^ + }}^\theta = 0.2}$ V
${{2.303RT} \over F} = 0.059V$
Explanation:
$ \begin{aligned} & \mathrm{H}_{2(\mathrm{~g})}+\mathrm{M}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{M}_{\text {(aq) }}^{+}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \\\\ & \mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {anode }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right] \times 1^2}{\left[\mathrm{M}^{+3}\right] 1} \\\\ & 0.1115=0.2-\frac{0.059}{2} \log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & 3=\log \frac{\left[\mathrm{M}^{+}\right]}{\left[\mathrm{M}^{+3}\right]} \\\\ & \therefore \mathrm{a}=3 \end{aligned} $
Consider the cell
$\mathrm{Pt(s)|{H_2}(g)\,(1\,atm)|{H^ + }\,(aq,[{H^ + }] = 1)||F{e^{3 + }}(aq),F{e^{2 + }}(aq)|Pt(s)}$
Given $\mathrm{E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.771\,V}$ and $\mathrm{E_{{H^ + }/1/2\,{H_2}}^o = 0\,V,\,T = 298\,K}$
If the potential of the cell is 0.712 V, the ratio of concentration of Fe$^{2+}$ to Fe$^{3+}$ is _____________ (Nearest integer)
Explanation:
At 298 K, a 1 litre solution containing 10 mmol of $\mathrm{C{r_2}O_7^{2 - }}$ and 100 mmol of $\mathrm{Cr^{3+}}$ shows a pH of 3.0.
Given : $\mathrm{C{r_2}O_7^{2 - } \to C{r^{3 + }}\,;\,E^\circ = 1.330}$V
and $\mathrm{{{2.303\,RT} \over F} = 0.059}$ V
The potential for the half cell reaction is $x\times10^{-3}$ V. The value of $x$ is __________
Explanation:
For lead storage battery pick the correct statements
A. During charging of battery, $\mathrm{PbSO}_{4}$ on anode is converted into $\mathrm{PbO}_{2}$
B. During charging of battery, $\mathrm{PbSO}_{4}$ on cathode is converted into $\mathrm{PbO}_{2}$
C. Lead storage battery consists of grid of lead packed with $\mathrm{PbO}_{2}$ as anode
D. Lead storage battery has $\sim 38 \%$ solution of sulphuric acid as an electrolyte
Choose the correct answer from the options given below:
The reaction
$\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$
occurs in which of the given galvanic cell.
The standard electrode potential of $\mathrm{M}^{+} / \mathrm{M}$ in aqueous solution does not depend on
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)
Assertion (A) : An aqueous solution of $\mathrm{KOH}$ when used for volumetric analysis, its concentration should be checked before the use.
Reason (R) : On aging, $\mathrm{KOH}$ solution absorbs atmospheric $\mathrm{CO}_{2}$.
In the light of the above statements, choose the correct answer from the options given below :
Which one of the following statements is correct for electrolysis of brine solution?
The standard electrode potential $\mathrm{(M^{3+}/M^{2+})}$ for V, Cr, Mn & Co are $-$0.26 V, $-$0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate $\mathrm{H_2}$ from a dilute acid are :
Choose the correct representation of conductometric titration of benzoic acid vs sodium hydroxide.
For a cell, $\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \,\mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \,\mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$
the cell potential is found to be $0.43 \mathrm{~V}$ at $298 \mathrm{~K}$. The magnitude of standard electrode potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is _________ $\times 10^{-2} \mathrm{~V}$.
[Given : $E_{A{g^ + }/Ag}^\Theta $ = 0.80 V and ${{2.303RT} \over F}$ = 0.06 V]
Explanation:
Cathode : $\left.[\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})\right] 2$
Cus(s) $+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\ & 0.43=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left(\frac{10^{-3}}{\left(10^{-2}\right)^{2}}\right) \\ & 0.43=E_{\text {cell }}^{0}-0.03 \log 10 \\ & \mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V} \\ & \mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} \\ & \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \mathrm{~V}=34 \times 10^{-2} \end{aligned} $
Resistance of a conductivity cell (cell constant $129 \mathrm{~m}^{-1}$) filled with $74.5 \,\mathrm{ppm}$ solution of $\mathrm{KCl}$ is $100 \,\Omega$ (labelled as solution 1). When the same cell is filled with $\mathrm{KCl}$ solution of $149 \,\mathrm{ppm}$, the resistance is $50 \,\Omega$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $\frac{\wedge_{1}}{\wedge_{2}}=x \times 10^{-3}$. The value of $x$ is __________. (Nearest integer)
Given, molar mass of $\mathrm{KCl}$ is $74.5 \mathrm{~g} \mathrm{~mol}^{-1}$.
Explanation:
$\mathrm{KCl}$ solution $1 \Rightarrow 74.5 \,\mathrm{ppm}, \mathrm{R}_{1}=100 \Omega$
$\mathrm{KCl}$ solution $2 \Rightarrow 149 \,\mathrm{ppm}, \mathrm{R}_{2}=50 \Omega$
$ \begin{aligned} &\text { Here, } \frac{p p m_{1}}{p p m_{2}}=\frac{M_{1}}{M_{2}}=\left(\frac{w_{1 / M_{0}}}{V} \times \frac{V}{w_{2 / M_{0}}}\right) \\ &\frac{\Lambda_{1}}{\Lambda_{2}}=\frac{k_{1} \times \frac{1000}{M_{1}}}{k_{2} \times \frac{1000}{M_{2}}} \\ &=\frac{k_{1}}{k_{2}} \times \frac{M_{1}}{M_{2}} \\ &=\frac{50}{100} \times 2 \\ &=\frac{\Lambda_{1}}{\Lambda_{2}}=1000 \times 10^{-3} \\ &=1000 \end{aligned} $
The amount of charge in $\mathrm{F}$ (Faraday) required to obtain one mole of iron from $\mathrm{Fe}_{3} \mathrm{O}_{4}$ is ___________. (Nearest Integer)
Explanation:
$ x=\frac{+8}{3} $
where x is oxidation state of Fe.
$ \mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O} $
Charge required $=\frac{8}{3} \times \mathrm{F}=\frac{8 \mathrm{~F}}{3} \simeq 3 \mathrm{~F}$
The spin-only magnetic moment value of M3+ ion (in gaseous state) from the pairs Cr3+ / Cr2+, Mn3+ / Mn2+, Fe3+ / Fe2+ and Co3+ / Co2+ that has negative standard electrode potential, is ____________ B.M. [Nearest integer]
Explanation:
$\operatorname{Cr}$ (III) $\Rightarrow d^{3}$
Number of unpaired electrons $=3$
$\mu=\sqrt{3(3+2)}=\sqrt{15} \simeq 4$ B.M.
The cell potential for $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{Sn}^{x+}\right| \mathrm{Sn}$ is $0.801 \mathrm{~V}$ at $298 \mathrm{~K}$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is ____________.
$\left(\right.$ Given $: \mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}^{\mathrm{o}}=-0.763 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{x+} \mid \mathrm{Sn}}^{\mathrm{o}}=+0.008 \mathrm{~V}$ and $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)$
Explanation:
$\mathrm{C}: \mathrm{Sn}^{+\mathrm{x}}+\mathrm{xe}^{-} \rightarrow \mathrm{Sn}$
$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\mathrm{Zn} \mid \mathrm{Zn}^{2+}}^{\circ}+\mathrm{E}_{\mathrm{Sn}^{+x} \mid \mathrm{Sn}}^{\circ}$
$\Rightarrow 0.763+0.008=0.771 \mathrm{~V}$
From the Nernst equation,
$\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^{\circ} \frac{-2.303 \,\mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}$
$0.801=0.771-\frac{0.06}{\mathrm{n}} \log 10^{-2}$
$0.03=\frac{0.06}{n} \times 2$
$\mathrm{n}=4$
The cell potential for the given cell at 298 K
Pt| H2 (g, 1 bar) | H+ (aq) || Cu2+ (aq) | Cu(s)
is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10$-$x M. The value of x is ___________.
(Given : $E_{C{u^{2 + }}/Cu}^\Theta $ = 0.34 V and ${{2.303\,RT} \over F}$ = 0.06 V)
Explanation:
$ \begin{aligned} &E=E_{\text {cell }}^{\circ}-\frac{0.06}{n} \log Q \\\\ &0.31=0.34-\frac{0.06}{2} \log \frac{10^{-6}}{C} \\\\ &\log \frac{10^{-6}}{C}=1 \\\\ &C=10^{-7} M \\\\ &x=7 \end{aligned} $
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced a STP is _____________ cm3. (Nearest integer)
[Given : Faraday constant F = 96500 C mol$-$1 at STP, molar volume of an ideal gas is 22.7 L mol$-$1]
Explanation:
$0.10 \times 2 \times 3600$ coulomb produces
$ \begin{aligned} & =\frac{\frac{3}{2} \times 0.1 \times 2 \times 3600}{2 \times 96500} \\\\ & =0.0056 \text { moles of gas } \end{aligned} $
Volume of gas produced $=0.0056 \times 22.7 \mathrm{~L}$
$ \begin{aligned} & \simeq 0.127 \mathrm{~L} \\\\ & =127 \mathrm{~mL} \end{aligned} $