1. a) calculate gibbs free energy for the following reaction at 25 degree celcius Au(s)+Ca(2+ 1 aq)----------------- Au(3+ aq 1M)+Ca(s)

EAu(3+)/Au=+1.50 V; E Ca(2+)/Ca=-2.87 V

b)Predict whether the reaction will be spontaneous or not

(i) The balanced redox reaction is as follows:
2Au (s) + 3Ca²⁺ (aq) → 2Au³⁺ (aq)  +  3Ca(s)

As we know Gibbs energy is given by
$\Delta G^0 = -nF{E^0}_{cell}$
where n is number of electron involved in reaction, E^​0_cell is emf of cell.

Reduction half Reaction : 3Ca²⁺ + 6e^- → 3Ca
Oxidation half reaction :  2Au + 6e^- ​→ 2Au³⁺

So, number of electrons involved in reaction is 6. 
E_cell = E_Right - E_Left = E_Ca²⁺/Ca - E_Au³⁺/Au
E_cell = -2.87 - 1.50
E_cell = -4.37 V
The concentration of all species is taken as unity.Therefore, E_cell = E⁰_cell = -4.37 V
Now put the value of E^​0_cell and n = 6 and F = 96487 C mol^-1

$$\begin{gathered} \Delta G^0 =- 6\times 96487\times (-4.37) \\ \Delta G^0 = 2,529,889.14 J mo{l}^{-1} \\ \Delta G^0 = 2529.889kJmo{l}^{-1} \end{gathered}$$

(b) The reaction is non-spotaneous as Gibbs energy of the reaction is positive.