Q. 1 g of a metal combines with 8.89 g of Br.Eq wt of metal is

Q.

When sodium acetate is added to aqueous solution of acetic acid, then why does the ph value increase?​

Q. If excess of $Zn$ is added to $1.0M$ solution of $CuS{O}_4$, find the concentration of $C{u}^{2+}$ ions at equilibrium.
Given: $E_{\left(Z{n}^{2+}|Zn\right)}^0=-0.76V,E_{\left(C{u}^{2+}Cu\right)}^0=-0.34V$

Q. Paragraph for below question
There are two types of concentration cells.
(i) Electrode concentration Cell: In this type of cell electrode has different concentration of electrode material in the same electrolyte.
Ex. Pt|H₂(P₁ atm)|HCl(cM)|H₂(P₂ atm)|Pt
(ii) Electrolyte Concentration Cell: In this type of cell, same electrode is immersed in the electrolyte having common species but different concentrations.
Ex. Pt|H₂(1 atm)|HCl(c₁M)||HCl(c₂M)|H₂ (1 atm)|Pt
Q. Calculate emf of the following concentration cell at 25°C.
Pt|H₂(1 atm)|HCl(10^–5 M)||CH₃COOH(0.1 M)|H₂(1 atm)|Pt
[Given: K_a (CH₃COOH) = 10^–5. Use $\frac{2.3030\mathrm{RT}}{F}=0.06$]
नीचे दिए गए प्रश्न के लिए अनुच्छेद
सांद्रता सेल दो प्रकार के होते हैं :
(i) इलेक्ट्रॉड सांद्रता सेल: इस प्रकार के सेल में, एक ही विद्युतअपघटय में स्थित इलेक्ट्रॉडों के पदार्थ की सांद्रता भिन्‍न होती है।
उदाहरण : Pt|H₂(P₁ atm)| HCl(cM)|H₂(P₂ atm)|Pt
(ii) विद्युतअपघट्य सांद्रता सेल: इस प्रकार के सेल में समान इलेक्ट्रॉड, समान स्‍पीशीज लेकिन भिन्‍न सांद्रताओं वाले विद्युत अपघट्य में‍ डूबे रहते हैं।
उदाहरण : Pt|H₂(1 atm)|HCl(c₁M)|| HCl(c₂M)|H₂ (1 atm)| Pt

प्रश्न. 25°C पर निम्नलिखित सांद्रता सेल के वि.वा.ब. (emf) की गणना कीजिए
Pt|H₂(1 atm)|HCl(10^–5 M)||CH₃COOH(0.1 M)|H₂(1 atm)|Pt
[दिया गया है: K_a(CH₃COOH) = 10^–5, प्रयोग करें : $\frac{2.3030\mathrm{RT}}{F}=0.06$]

1. 0.12 V
2. 0.18 V
3. 0.03 V
4. 0.06 V

Q. Paragraph for below question
There are two types of concentration cells.
(i) Electrode concentration Cell: In this type of cell electrode has different concentration of electrode material in the same electrolyte.
Ex. Pt|H₂(P₁ atm)|HCl(cM)|H₂(P₂ atm)|Pt
(ii) Electrolyte Concentration Cell: In this type of cell, same electrode is immersed in the electrolyte having common species but different concentrations.
Ex. Pt|H₂(1 atm)|HCl(c₁M)||HCl(c₂M)|H₂ (1 atm)|Pt
Q. EMF of the following electrochemical cell Pt|H₂(1 atm)|HA(0.1 M)||HCl(0.1 M)|H₂(1 atm)|Pt is 0.18 V at 25°C. Then calculate K_a of weak acid HA. $\left(\mathrm{Use}\frac{\mathrm{RT}\times 2.303}{F}=0.06\right]$
नीचे दिए गए प्रश्न के लिए अनुच्छेद
सांद्रता सेल दो प्रकार के होते हैं :
(i) इलेक्ट्रॉड सांद्रता सेल: इस प्रकार के सेल में, एक ही विद्युतअपघटय में स्थित इलेक्ट्रॉडों के पदार्थ की सांद्रता भिन्‍न होती है।
उदाहरण : Pt|H₂(P₁ atm)| HCl(cM)|H₂(P₂ atm)|Pt
(ii) विद्युतअपघट्य सांद्रता सेल: इस प्रकार के सेल में समान इलेक्ट्रॉड, समान स्‍पीशीज लेकिन भिन्‍न सांद्रताओं वाले विद्युत अपघट्य में‍ डूबे रहते हैं।
उदाहरण : Pt|H₂(1 atm)|HCl(c₁M)|| HCl(c₂M)|H₂ (1 atm)| Pt

प्रश्न. 25°C पर विद्युत रासायनिक सेल Pt|H₂(1 atm)|HA(0.1 M)||HCl(0.1 M)|H₂(1 atm)|Pt का वि.वा.ब. (EMF) 0.18 V है। तब दुर्बल अम्ल HA के K_a की गणना कीजिए।

1. 2 × 10^–5
2. 5 × 10^–6
3. 1 × 10^–8
4. 1 × 10^–7

Q. The standard reduction potential at STP $({25}^{\circ }C$ and 1 atm) of the following half cells are found as below.
$P{b}_{aq}^{2+}+2e^{-}\to P{b}_{\left(s\right)}{E}^0=-0.121VPb{O}_{2\left(s\right)}+4H_{\left(aq\right)}^{+}+S{O}_{4\left(aq\right)}^{2-}+2e^{-}\to PbS{O}_{4\left(s\right)}+2H_{2}O{E}^0=1.71VP{b}_{\left(aq\right)}^{4+}+2e^{-}\to P{b}_{\left(aq\right)}^{2+}{E}^0=1.689V$
For the cell $PbS{O}_{4\left(s\right)}+2e^{-}\to P{b}_{\left(s\right)}+S{O}_{4\left(aq\right)}^{2-}$,
The $E^0$ is missing but it is found that in an aqueous solution of $H_{2}S{O}_4$ having a pH=1.7, when $PbS{O}_{4\left(s\right)}$ is put, then at saturated condition
Solubility of $PbS{O}_4=5\times {10}^{-5}Mol/L$
$(Take\frac{2.303RT}{F}=0.06andlog2=0.3)$

List-I contains questions and List-II contains their answers.
$\begin{array}{cc}\text{List-I}& \text{List-II}\\ \text{What is the}{E}^0\text{(in V) for the half-cell}& \\ \text{(I)}PbS{O}_{4\left(s\right)}+2e^{-}\to P{b}_{\left(s\right)}+S{O}_{4\left(aq\right)}^{2-}?& \text{(P) -0.168}\\ \text{What is the}{E}_{cell}^0\text{(in V) of a lead storage battery? It has}& \\ \text{(II) the cell reaction as}& \\ P{b}_{\left(s\right)}+Pb{O}_{2\left(s\right)}+2H_{2}S{O}_{4_{\left(aq\right)}}\to 2PbS{O}_{4\left(s\right)}+2H_2{O}_{\left(l\right)}& \text{(Q) 0.7}\\ \text{What is the}{E}^0\text{(in V) for the half cell}& \\ \text{(III)}Pb{O}_{2\left(s\right)}+4H_{\left(aq\right)}^{+}+4e^{-}\to P{b}_{\left(s\right)}+2H_2{O}_{\left(I\right)}& \text{(R) 2.0}\\ K_{sp}\text{is the equilibrium constant for the reaction}& \\ \text{(IV)}Pb{O}_{2\left(s\right)}+4H_{\left(aq\right)}^{+}\rightleftharpoons P{b}_{\left(aq\right)}^{4+}+2H_2{O}_{\left(l\right)}& \text{(S) -0.068}\\ \text{What is the value of}{E}^o\text{(in V)}?& \\ & \text{(T) -0.31}\\ & \text{(U) 2.02}\end{array}$
Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?

1. (II), (R) and (IV), (S)
2. (II), (U) and (IV), (S)
3. (II), (R) and (IV), (P)
4. (II), (U) and (IV), (P)

Q. The pH of a solution at 25 degree Celsius containing 0.20 M sodium acetate and 0.06 M acetic acid is (pKa for CH3COOH = 4.74 and log 3 = 0.477) a) 4.36 b) 5.26 c) 5.84 d) 6.32

Q. The pressure of $H_2$ required to make the potential of $H_2$ electrode zero in pure water at 298 K is

1. ${10}^{-10}atm$
2. ${10}^{-4}atm$
3. ${10}^{-14}atm$
4. ${10}^{-12}atm$

Q. The cell reaction for the given cell is spontaneous if $Pt,C{l}_{2_{\left(P_1\right)}}\left(g\right)|C{l}^{-}\left(1.0M\right)||C{l}^{-}\left(2M\right)|C{l}_{2_{\left(P_2\right)}}\left(g\right),Pt$

1. $P_1>P_2$
2. $2P_1<P_2$
3. $P_1=P_2$
4. $2P_2>P_1$

Q. The standard EMF of a galvanic cell can be calculated from:

1. the amount of metal in the anode
2. the size of the electrode
3. the $E^o$ values of the half-cells
4. the pH of the solution

Q. Consider the following cell reaction:
$2Fe\left(s\right)+O_2\left(g\right)+4H^{+}\left(aq\right)\to 2F{e}^{2+}\left(aq\right)+2H_{2}O\left(l\right);E^{\circ }=1.67V$
At $\left[F{e}^{2+}\right]={10}^{-3}M,P_{\left(O_2\right)}=0.1atm$ and pH = 3, the cell potential at $25{}^{\circ }C$ is
(given: $\frac{2.303RT}{F}=0.0591$)

Q. Mr.Nernst had an apprentice named Electra. She is hard-working but is not very good with Chemistry. She needs a lot of help from you to pass as an apprentice. Please help the damsel in distress!

One day Mr.Nernst told her to find the EMF of the following cell using experiments at ${25}^{\circ }C$
$Mg\left(s\right)+2A{g}^{+}\left(0.0001M\right)\to M{g}^{2+}\left(0.130M\right)+2Ag\left(s\right)$

Electra reported the value to be +2.6V. Mr.Nernst told her to recheck the value using his theory, knowing Standard EMF = 3.17 V.
What was the right value? Round off the answer to nearest integer.

Type your answer here. ___

Q.

Electra has trouble finding the correct Nernst equation for the cell $Zn\left(s\right)|Z{n}^{2+}||C{u}^{2+}|Cu\left(s\right)at{25}^{\circ }C$
Can you help her?

1. $E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}\right)$

2. $E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[C{u}^{2+}\right]}{\left[Z{n}^{2+}\right]}\right)$

3. $E_{cell}=E^{\circ }-0.0296log\left(\frac{Zn}{Cu}\right)$

4. $E_{cell}=E^{\circ }-0.0296log\left(\frac{Cu}{Zn}\right)$

Q. In the electrochemical cell:
$Zn|ZnS{O}_4\left(0.01M\right)||CuS{O}_4\left(1.0M\right)Cu$, the emf of this Daniell cell is $E_1$. When the concentration of $ZnS{O}_4$ is changed to 1.0 M and that of $CuS{O}_4$ changed to 0.01 M, the emf changes to $E_2$. From the followings, which one is the relationship between $E_1$ and $E_2$? (Given, RT/F = 0.059)

1. $E_1<E_2$
2. $E_1>E_2$
3. $E_2=-E_1$
4. $E_1=E_2$

Q. What is the physical significance of R and why is always Cp greater than Cp?

Q. The standard reduction potential at STP $({25}^{\circ }C$ and 1 atm) of the following half cells are found as below.
$P{b}_{aq}^{2+}+2e^{-}\to P{b}_{\left(s\right)}{E}^0=-0.121VPb{O}_{2\left(s\right)}+4H_{\left(aq\right)}^{+}+S{O}_{4\left(aq\right)}^{2-}+2e^{-}\to PbS{O}_{4\left(s\right)}+2H_{2}O{E}^0=1.71VP{b}_{\left(aq\right)}^{4+}+2e^{-}\to P{b}_{\left(aq\right)}^{2+}{E}^0=1.689V$
For the cell $PbS{O}_{4\left(s\right)}+2e^{-}\to P{b}_{\left(s\right)}+S{O}_{4\left(aq\right)}^{2-}$,
The $E^0$ is missing but it is found that in an aqueous solution of $H_{2}S{O}_4$ having a pH=1.7, when $PbS{O}_{4\left(s\right)}$ is put, then at saturated condition
Solubility of $PbS{O}_4=5\times {10}^{-5}Mol/L$
$(Take\frac{2.303RT}{F}=0.06andlog2=0.3)$

List-I contains questions and List-II contains their answers.
$\begin{array}{cc}\text{List-I}& \text{List-II}\\ \text{What is the}{E}^0\text{(in V) for the half-cell}& \\ \text{(I)}PbS{O}_{4\left(s\right)}+2e^{-}\to P{b}_{\left(s\right)}+S{O}_{4\left(aq\right)}^{2-}?& \text{(P) -0.168}\\ \text{What is the}{E}_{cell}^0\text{(in V) of a lead storage battery? It has}& \\ \text{(II) the cell reaction as}& \\ P{b}_{\left(s\right)}+Pb{O}_{2\left(s\right)}+2H_{2}S{O}_{4_{\left(aq\right)}}\to 2PbS{O}_{4\left(s\right)}+2H_2{O}_{\left(l\right)}& \text{(Q) 0.7}\\ \text{What is the}{E}^0\text{(in V) for the half cell}& \\ \text{(III)}Pb{O}_{2\left(s\right)}+4H_{\left(aq\right)}^{+}+4e^{-}\to P{b}_{\left(s\right)}+2H_2{O}_{\left(I\right)}& \text{(R) 2.0}\\ K_{sp}\text{is the equilibrium constant for the reaction}& \\ \text{(IV)}Pb{O}_{2\left(s\right)}+4H_{\left(aq\right)}^{+}\rightleftharpoons P{b}_{\left(aq\right)}^{4+}+2H_2{O}_{\left(l\right)}& \text{(S) -0.068}\\ \text{What is the value of}{E}^o\text{(in V)}?& \\ & \text{(T) -0.31}\\ & \text{(U) 2.02}\end{array}$
Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?

1. (I), (S) and (III), (P)
2. (I), (T) and (III), (Q)
   
3. (I), (Q) and (III), (R)
4. (I), (U) and (III), (R)

Q. A hydrogen electrode placed in a solution containing sodium acetate and acetic acid in the concentration ratios of $x:y$ and $y:x$ has electrode potential values $E_1$ and $E_2$ volts, respectively, at ${25}^{\circ }\text{C}$. The $p{K}_a$ value of acetic acid is:

1. $\frac{-(E_1+E_2)}{2\times 0.059}$
2. $\frac{(E_1+E_2)}{2\times 0.059}$
3. $\frac{(E_2-E_1)}{2\times 0.059}$
4. $\frac{-(E_1+E_2)}{0.059}$

Q. <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> The correct order of following 3d metal oxides, according to their oxidation number is:
(a) $Cr{O}_3$
(b) $F{e}_2{O}_3$
(c) $Mn{O}_2$
(d) $V_2{O}_5$
(e) $C{u}_{2}O$


JEE MAIN 2021

Q. An electrochemical cell is constructed by immersing a piece of copper wire in 50 ml of 0.1 M $CuS{O}_4$ solution and zinc strip in 50 ml of 0.1 M $ZnS{O}_4$ solution
[ $E_{C{u}^{2+}/Cu}^0$= 0.34 V, $E_{Z{n}^{2+}/Zn}^0$= -0.76 V ]

The emf of the cell increases when small amount of concentrated $N{H}_3$ is added to:

1. $ZnS{O}_4$ solution
2. $CuS{O}_4$ solution
3. Both $ZnS{O}_4$ and $CuS{O}_4$
4. Can't say

Q. Consider the following cell
$Pt|H_2\left(P_{1}atm\right)|H^{+}\left(M_1\right)||H^{+}\left(M_2\right)|H_2\left(P_{2}atm\right)|Pt$
where $P_1$ and $P_2$ are pressures. $M_1$ and $M_2$ are molarities .
What will be the emf of cell at ${25}^{\circ }C$ If $P_1=P_2$ and $M_1$ is 50% higher than $M_2$ ?
Repeat the magnitude of emf by multiplying with 10000.
[Take : $\frac{2.303RT}{F}=0.06\text{and log}3=0.48,\text{log}2=0.3]$

Q.

Electra has trouble finding the correct Nernst equation for the cell $Zn\left(s\right)|Z{n}^{2+}||C{u}^{2+}|Cu\left(s\right)at{25}^{\circ }C$
Can you help her?

1. $E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}\right)$

2. $E_{cell}=E^{\circ }-0.0296log\left(\frac{\left[C{u}^{2+}\right]}{\left[Z{n}^{2+}\right]}\right)$

3. $E_{cell}=E^{\circ }-0.0296log\left(\frac{Zn}{Cu}\right)$

4. $E_{cell}=E^{\circ }-0.0296log\left(\frac{Cu}{Zn}\right)$

Q. For the following electrochemical cell at 298 K,
$Pt\left(s\right)|H_2(g,1bar)|H^{+}(aq,1M)||M^{4+}\left(aq\right),M^{2+}\left(aq\right)|Pt\left(s\right)$
$E_{cell}=0.092Vwhen\frac{[M^{2+}+(aq\left)\right]}{\left[M^{4+}\right(aq\left)\right]}={10}^x$
$Given:E_{M^{4+}/M^{2+}}^{\circ }=0.151V;2.303\frac{RT}{F}=0.059V$
The value of x is

1. -2
2. -1
3. 1
4. 2

Q. Which of the following cells have a cell potential which is the same as its standard value?

1. $Zn\left(s\right)|Z{n}^{2+}\left(aq\right)\left(0.02M\right)||H_3{O}^{+}\left(aq\right)\left(0.1M\right)|H_2\left(g\right)\left(0.5atm\right),Pt$
2. $Cu\left(s\right)|C{u}^{2+}\left(aq\right)\left(0.25M\right)||A{g}^{+}\left(aq\right)\left(0.5M\right)|Ag$
3. $Cd\left(s\right)|C{d}^{2+}\left(aq\right)\left(0.01M\right)||H^{+}(pH=1)|H_2\left(g\right)\left(1atm\right),Pt$
4. $Zn\left(s\right)|Z{n}^{2+}\left(aq\right)\left(0.1M\right)||H^{+}(pH=1)|H_2\left(g\right)\left(1atm\right),Pt$

Q.

The Edison storage cell is represented as :
Fe(s)/FeO(s)/KOH(aq)/Ni₂O₃ (s)/Ni(s)
The half-cell reactions are:
$N{i}_2{O}_3\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to 2NiO\left(s\right)+2O{H}^{-};E^{\circ }+0.40$
$FeO\left(s\right)+H_{2}O\left(l\right)+2e^{-}\to Fe\left(s\right)+2O{H}^{-};E^{\circ }=-0.87$,
What is the maximum amount of electrical energy that can be obtained from one mole of Ni₂O₃ ?

1. 344.6 KJ

2. 87.9 KJ

3. 2478.8 KJ

4. 245.11 KJ

Q.

Calculate the emf of the cell

Pt, $H_2\left(1.0atm\right)$ | $C{H}_{3}COOH$ (0.1M) || $N{H}_3$(aq, 0.001M) | $H_2$(1.0 atm), Pt Ka($C{H}_{3}COOH$) = ${10}^{-5}$, Kb(NH3) = ${10}^{-5}$

1. +0.368 V

2. -0.413 V

3. -0.233 V

4. +0.567 V

Q.

The e.m.f. of the cell at ${25}^{\circ }C$
$\frac{Cu}{C{u}^{2+}\left(0.01M\right)}||\frac{A{g}^{+}\left(0.1M\right)}{Ag}$
is $[E^0\left(\frac{C{u}^{2+}}{Cu}\right)=0.34Vand{E}^0\left(\frac{Ag+}{Ag}\right)=0.80V]$

1. 0.46V

2. 1.14V

3. 0.43V

4. 1.11V

Q. Mr.Nernst had an apprentice named Electra. She is hard-working but is not very good with Chemistry. She needs a lot of help from you to pass as an apprentice. Please help the damsel in distress!

One day Mr.Nernst told her to find the EMF of the following cell using experiments at ${25}^{\circ }C$
$Mg\left(s\right)+2A{g}^{+}\left(0.0001M\right)\to M{g}^{2+}\left(0.130M\right)+2Ag\left(s\right)$

Electra reported the value to be +2.6V. Mr.Nernst told her to recheck the value using his theory, knowing Standard EMF = 3.17 V.
What was the right value? Round off the answer to nearest integer.

Type your answer here. ___