Gibb's Energy and Nernst Equation

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. 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. -152 kJ mol^-1
2. -322 kJ mol^-1
3. -161 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. 2.14V

2. 4.28V

3. 8.56V

4. 6.42V

Q. The following reaction describes the rusting of iron.
$4Fe+3O_2\to 4F{e}^{3+}+6O^{2-}$

1. This is an example of a redox reaction.
2. Metallic iron is reduced to $F{e}^{3+}$.
3. $F{e}^{3+}$ is an oxidising agent.
4. Metallic iron is a reducing agent.

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. − 76
2. – 322
3. – 122
4. – 176

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