Question

The rusting of iron takes place as follows:
$\
  2{H^ \oplus } + 2{e^ - } + \dfrac{1}{2}{O_2} \to {{\text{H}}_2}{\text{O}}\left( l \right);{E^ \odot } = + 1.23V \\
  F{e^{2 + }} + 2{e^ - } \to Fe\left( s \right);{E^ \odot } = - 044V \\
\ $
Calculate $\Delta {G^ \odot }$ for the net process.
A.$ - 322{\text{ kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
B.$ - 161{\text{ kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
C.$ - 125{\text{ kJmo}}{{\text{l}}^{{\text{ - 1}}}}$
D.$ - 76{\text{ kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

Answer 1

Hint: Standard cell potential of both reactions is given. Use this formula to calculate Gibbs energy change-
$\Delta {G^ \odot } = - nF{E^ \odot }$ where n=moles of electrons from balanced Redox reaction, F is faraday constant whose value is $96458{\text{ C/mol}}$ and ${E^ \odot }$ is the standard cell potential .Then apply $\Delta G_{net}^ \odot = \Delta G_1^ \odot + \Delta G_2^ \odot $to calculate the net value.

Step-by-Step Explanation-The Given reactions are-
At Cathode:
 $\
  2{H^ \oplus } + 2{e^ - } + \dfrac{1}{2}{O_2} \to {{\text{H}}_2}{\text{O}}\left( l \right);{E^ \odot } = + 1.23V \\
    \\
\ $
Then n=$2$ .Now using formula-
$ \Rightarrow $ $\Delta {G^ \odot } = - nF{E^ \odot }$ Where n=moles of electrons from balanced Redox reaction, F is faraday constant whose value is $96458{\text{ C/mol}}$ and ${E^ \odot }$ is the standard cell potential
On putting the given values we get,
$ \Rightarrow \Delta G_1^ \odot = - 2 \times F \times \left( {1.23} \right)$
On solving we get,
$ \Rightarrow \Delta G_1^ \odot = - 2.46F$ --- (i)
Now at Anode:
$F{e^{2 + }} + 2{e^ - } \to Fe\left( s \right);{E^ \odot } = - 044V$
Then n=$2$.Now using formula-
$ \Rightarrow $ $\Delta {G^ \odot } = - nF{E^ \odot }$ Where n=moles of electrons from balanced Redox reaction, F is faraday constant whose value is $96458{\text{ C/mol}}$ and ${E^ \odot }$ is the standard cell potential
On putting the given values we get,
$ \Rightarrow \Delta G_2^ \odot = - 2 \times F \times \left( {0.44} \right)$
On solving we get,
$ \Rightarrow \Delta G_2^ \odot = - 0.88F$ --- (ii)
Now on applying formula
$ \Rightarrow \Delta G_{net}^ \odot = \Delta G_1^ \odot + \Delta G_2^ \odot $
On putting the values from eq. (i) and (ii) in this formula we get,
$ \Rightarrow \Delta G_{net}^ \odot = \left[ { - 2.46F} \right] + \left[ { - 0.88F} \right]$
On simplifying we get,
$ \Rightarrow \Delta G_{net}^ \odot = - 3.34F$
And we know the value of Faraday constant, so on putting the value we get,
$ \Rightarrow \Delta G_{net}^ \odot = - 3.34 \times 96458$${\text{Jmo}}{{\text{l}}^{{\text{ - 1}}}}$
$ \Rightarrow \Delta G_{net}^ \odot = - 322169.72$ ${\text{Jmo}}{{\text{l}}^{{\text{ - 1}}}}$
We know that $1{\text{KJ = 1000J}}$
Then$\Delta G_{net}^ \odot = - 322169.72 \times 1000{\text{ KJmo}}{{\text{l}}^{{\text{ - 1}}}}$
\[ \Rightarrow \Delta G_{net}^ \odot = - 322.169{\text{ KJmo}}{{\text{l}}^{{\text{ - 1}}}}\]

Hence correct option is A.

Note: $\Delta {G^ \odot }$ is Gibbs energy change for a system under standard conditions while $\Delta G$ is Gibbs free energy for a system. $\Delta {G^ \odot }$ is also given as –
$ \Rightarrow \Delta {G^ \odot } = - RT\ln K$
Where $R = 8.314{\text{ Jmol}}{{\text{C}}^{ - 1}}$ is gas constant, T=Temperature and K is equilibrium constant of a reaction.
Gibbs free energy is given as-$\Delta G = \Delta H - T\Delta S$ where $\Delta H$ is change in enthalpy, $\Delta S$ is change in entropy and T is the temperature.