Question

Calculate the standard Gibbs energy (Δ_rG°) of the following reaction at 25°C:

\[\ce{Au_{(s)} + Ca^2 + (1M) ->Au^3 + (1M) + Ca_{(s)}}\] 

`E_(Au^(3+)"/"Au)^° = +1.5  V, E_(Ca^(2+)"/"Ca)^° = -2.87  V`

Predict whether the reaction will be spontaneous or not at 25 °C. [1 F = 96500 C mol⁻¹]

Answer 1

Given: `E_(Au^(3+)"/"Au)^° = +1.5  V`,

`E_(Ca^(2+)"/"Ca)^° = -2.87  V`

\[\ce{Au_{(s)} + Ca^2 + (1M) ->Au^3 + (1M) + Ca_{(s)}}\] 

`E_"cell"^° = E_"cathode"^° − E_"anode"^°`

= (−2.87 V) − (1.5 V)

= −4.37 V

In this reaction, the oxidation half-reaction shows the loss of 3 electrons (\[\ce{Au -> Au3+ + 3e-}\]) and the reduction half-reaction shows the gain of 2 electrons (\[\ce{Ca2+ + 2e -> Ca}\]).

To balance the electrons, the least common multiple (LCM) of 2 and 3 is 6, so the overall reaction involves the transfer of 6 electrons.

The standard Gibbs free energy change is calculated using:

ΔG° = `-  nFE°`

ΔG° = −(6) (96500 C/mol) (−4.37V)

Since 1 V = 1 J C⁻¹, the coulomb units cancel, leaving the result in joules.

ΔG° = −6 × 96500 × (−4.37) 

= +2530230 J/mol 

= +2530.23 KJ/mol

Since the value of ΔG° is positive, the reaction is nonspontaneous at 25°C.