Nersnt Equation

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

1. $5\times {10}^{-9}$
2. $5\times {10}^{-25}$
3. $5\times {10}^{-17}$
4. $5\times {10}^{-38}$

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.

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. Electra made a concentrated solution of $ZnS{O}_4$ and calculated the EMF of $\frac{Z{n}^{2+}}{Zn}$ half-cell using experiments. This time she was certain that the experiment was right. She still tried to verify it using her master Nernst's equation. She found it was inaccurate! She concluded that Nernst equation must be wrong or inapplicable in this case.
Her conclusion is right.

1. True
2. Flase

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.

The emf of the cell $Ag|AgI|KI\left(0.05M\right)||AgN{O}_3\left(0.05M\right)Ag$ is $0.788V$. Calculate the solubility product of $AgI$.

1. 1.10 x 10^-16
2. 8 x 10^-11

3. 25 X 10^-13

4. 6.6 x 10^-19

Q. Let's help Electra calculate the EMF of the cell $Zn\left(s\right)|Z{n}^{2+}\left(0.024M\right)||Z{n}^{2+}\left(2.4M\right)|Zn\left(s\right)$. Round off the answer to 2 decimal places.

1. 0.07 V
2. 0.06 V
3. 0.08 V
4. 0.1 V

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

1. $5\times {10}^{-17}$
2. $5\times {10}^{-38}$
3. $5\times {10}^{-9}$
4. $5\times {10}^{-25}$

Q. The hydrogen electrode is dipped in a solution of $pH=3$ at ${25}^{\circ }C$. The potential of the cell would be (the value of $2.303\frac{RT}{F}$is $0.059V$

1. 0.087 V
2. 0.059 V
3. 0.177 V
4. – 0.177 V

Q. The hydrogen electrode is dipped in a solution of The potential of the cell would be (the value of 2.303RT/F is 0.059 V)

1. 0.177 V

2. -0.177 V

3. 0.087 V

4. 0.059 V

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. 2
3. 1
4. -1