The EMF of the standard weston cadmium cell
`Cd(12.5%)` in `Hg|3CdSO_(4),8H_(2)O` (solid) | satd. Soln of `CdSO_(4)||Hg_(2)SO_(4)(s)|Hg` is
`1.0180` volts at `25^(@)C` and the temperature coefficient of the cell, `((delE)/(delT))_(P)=-4.0xx10^(-5)//"degree"`, calculate `triangleG` `triangleH` and `triangleS` for the reactionn in the cell when `n=2`

Explanation

To solve the problem, we will calculate the Gibbs free energy change (ΔG), enthalpy change (ΔH), and entropy change (ΔS) for the reaction in the standard Weston cadmium cell.

### Step 1: Calculate ΔG
The formula for Gibbs free energy change is given by:
\[ \Delta G = -nFE \]
where:
- \( n \) = number of moles of electrons transferred (given as 2)
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E \) = EMF of the cell (given as \( 1.0180 \, \text{V} \))

Substituting the values:
\[ \Delta G = -2 \times 96500 \times 1.0180 \]
Calculating this gives:
\[ \Delta G = -196.474 \, \text{kJ} \]

### Step 2: Calculate ΔS
The formula for the change in entropy is given by:
\[ \Delta S = nF \left( \frac{dE}{dT} \right)_{P} \]
where:
- \( \frac{dE}{dT} \) = temperature coefficient (given as \( -4.0 \times 10^{-5} \, \text{V/K} \))

Substituting the values:
\[ \Delta S = 2 \times 96500 \times (-4.0 \times 10^{-5}) \]
Calculating this gives:
\[ \Delta S = -7.72 \, \text{J/K} \]

### Step 3: Calculate ΔH
Using the relationship between ΔG, ΔH, and ΔS:
\[ \Delta G = \Delta H - T\Delta S \]
Rearranging gives:
\[ \Delta H = \Delta G + T\Delta S \]
Where:
- \( T \) = temperature in Kelvin (25°C = 298 K)

Substituting the values:
\[ \Delta H = -196.474 + 298 \times (-7.72 \times 10^{-3}) \quad \text{(converting J to kJ)} \]
Calculating this gives:
\[ \Delta H = -196.474 + (-2.30376) = -198.774 \, \text{kJ} \]

### Summary of Results
- \( \Delta G = -196.474 \, \text{kJ} \)
- \( \Delta H = -198.774 \, \text{kJ} \)
- \( \Delta S = -7.72 \, \text{J/K} \)