Gibb's Energy and Nernst Equation

Q. The electrochemical cell shown below is a concentration cell.$M|M^{2+}$ (Saturated solution of a sparingly soluble salt, $M{X}_2\left)||M^{2+}\right(0.001mold{m}^{-3})M.$ The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of the cell at 298K is 0.059

The value of $\Delta G\left(kJmo{l}^{-1}\right)$ for the given cell is: (take$1F=96500Cmo{l}^{-1})$

1. $5.7$
2. $11.4$
3. $-5.7$
4. $-11.4$

Q. The half-cell reactions for rusting of iron are,
$2H^{+}+\frac{1}{2}{O}_2+2e^{-}\to 2H_{2}O,E^{\circ }=+1.23V,$
$F{e}^{2+}+2e^{-}\to Fe\left(s\right);E^{\circ }=-0.44V.$
$\Delta G^{\circ }\left(inkJ\right)$ for the reaction is :

1. – 322
2. – 122
3. – 176
4. − 76

Q. Calculate standard free energy change for the reaction $\frac{1}{2}Cu\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\rightleftharpoons \frac{1}{2}C{u}^{2+}+C{l}^{-}$taking place at in a cell whose standard e.m.f. is $1.02volts$

1. – 98430 J
2. 98430 J
3. 96500 J
4. – 49215 J

Q.

Given standard electrode potentials
$F{e}^{++}+2e^{-}\to Fe;E^{\circ }=-0.440V$
$F{e}^{+++}+3e^{-}\to Fe;E^{\circ }=-0.036V$
The standard electrode potential $\left(E^{\circ }\right)$ for $F{e}^{+++}+e^{-}\to F{e}^{++}$ is

1. – 0.476 V
2. – 0.404 V
3. + 0.404 V
4. + 0.772 V

Q. The rusting of iron takes place as follows $2H^{+}+2e^{-}+\frac{1}{2}{O}_2\to H_{2}O\left(l\right);E^{\circ }=+1.23V$ $F{e}^{2+}+2e^{-}\to Fe\left(s\right);E^{\circ }=-0.44V$ Calculate $\Delta G^{\circ }$ for the net process

1. -322 kJ mol^-1
2. -161 kJ mol^-1
3. -152 kJ mol^-1
4. -76 kJ mol^-1

Q.

On the basis of information available from the reaction:
$\frac{4}{3}Al+O_2\to \frac{2}{3}A{l}_2{O}_3\Delta G=-827KJ/mol$ of $O_2$. The minimum e.m.f. required to carry out
electrolysis of $A{l}_2{O}_3$ is

1. 4.28V

2. 2.14V

3. 6.42V

4. 8.56V

Q. For the following cell:
$Zn\left(s\right)|ZnS{O}_4\left(aq\right)||CuS{O}_4\left(aq\right)\left)|Cu\right(s)$
When the concentration of $Z{n}^{2+}$ is 10 times the concentration of $C{u}^{2+}$, the expression for $\Delta G\left(inJmo{l}^{-1}\right)$ is
[F is Faraday constant; R Is gas constant; T is temperature ; $E^{\circ }\left(cell\right)=1.1v]$

1. 2.303 RT - 2.2 F
2. 1.1 F
3. -2.2 F
4. 2.303 RT + 1.1 F

Q. Cool! Electra has mastered Equilibrium constants and has moved on to Gibb's energy.
Electra was thinking, "When we calculate the Gibb's energy for a SHE using electrode potential, we get 0.
So is it right to say that this half-cell ‘has no change in Gibb's energy’ ?"
What will you tell her?

1. It's true Electra
2. Gibb’s energy cannot be applied here. Foolish Electra.
3. Half-cells don’t have electrode potential.
4. No way, that's wrong

Q.

Electra is almost bored now. Stupid Nernst making her solve so many problems.

But wait, this is her last one. If she completes this, she is through as an apprentice!
Quick. Tell her what is the Gibb's energy of a Daniel Cell.
Use $F=96487Cmo{l}^{-1}$

1. $21.227Jmo{l}^{-1}$
2. $21.227kJmo{l}^{-1}$
3. $21.227mJmo{l}^{-1}$
4. $21.227mo{l}^{-1}$

Q. Electra - "Hmm, this looks tricky"

At ${25}^{\circ }C$,
$C{r}^{3+}+e^{-}\to C{r}^{2+},given{E}^0=-0.424VC{r}^{2+}+2e^{-}\to Cr\left(s\right),given{E}_0=-0.9VFind{E}^{0}at{25}^{\circ }CforC{r}^{3+}\left(aq\right)+3e^{-}\to Cr\left(s\right)$

Electra was completely stumped. Seems like you need a way to add electrode potentials. Please tell her the answer.

1. 0.741 V
2. -1.324 V
3. - 0.741 V

4. -0.476 V