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.$- 76{\text{ kJmo}}{{\text{l}}^{{\text{ - 1}}}}$

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.