There are two principal types of electrochemical cells. A galvanic cell is an electrochemical cell that produces electricity as a result of spontaneous reaction occurring inside it. An electrolytic cell is an electrochemical cell in which a non-spontaneous reaction is driven by an external source of current. any redox reaction may be expressed in terms of two half reactions which are conceptual reactions showing the lowss and gain of electrons. each half reaction has a difinite value of standard electrode potential. the overall reaction is represented by a universally accepted method. knowing the standard electrode potential of the half reactions, the standard EMF of the cell can be calculted. the standard EMF further helps in the calculation of free energy change, equilibrium constant of the cell reaction as well as parameters like solublity products of a sparingly soluble salt. a cell can also be set up in which the two electrodes may be of the same (type, e.g., both may be hydrogen electrodes but the concentration of `H^(+)` ions in the two solutions may be different. Such cells are called concentration cells.
Q. The reaction
`(1)/(2)H_(2)(g)+AgCl(s)toH^(+)(aq)+Cl^(-)(aq)+Ag(s)`
occurs in the galvanic cell

Options

A. `Ag|AgCl(s)|KCl(aq)|AgNO_(3)(aq)|Ag`
B. `Pt|H_(2)(g)|HCl(aq)|AgNO_(3)(aq)|Ag`
C. `Pt|H_(2)(g)|HCl(aq)|AgCl(s)|Ag`
D. `Pt|H_(2)(g)|KCl(aq)|AgCl(s)|Ag`

Explanation

To solve the problem regarding the galvanic cell reaction given by:

\[ \frac{1}{2} H_2(g) + AgCl(s) \rightarrow H^+(aq) + Cl^-(aq) + Ag(s) \]

we will follow these steps:

### Step 1: Identify the half-reactions
The overall reaction can be broken down into two half-reactions: one for the oxidation and one for the reduction.

1. **Oxidation half-reaction**:
\[ \frac{1}{2} H_2(g) \rightarrow H^+(aq) + e^- \]

2. **Reduction half-reaction**:
\[ Ag^+(aq) + e^- \rightarrow Ag(s) \quad \text{(from AgCl)} \]

### Step 2: Write the standard electrode potentials
The standard electrode potentials (E°) for the half-reactions can be looked up in standard tables:

- For the oxidation of hydrogen:
\[ E^\circ_{H^+/H_2} = 0.00 \, \text{V} \]

- For the reduction of silver:
\[ E^\circ_{Ag^+/Ag} \approx +0.80 \, \text{V} \]

### Step 3: Calculate the standard EMF of the cell
The standard EMF (E°cell) of the galvanic cell can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

In this case:
- Cathode (reduction): Silver electrode
- Anode (oxidation): Hydrogen electrode

Thus:
\[ E^\circ_{cell} = E^\circ_{Ag^+/Ag} - E^\circ_{H^+/H_2} \]
\[ E^\circ_{cell} = 0.80 \, \text{V} - 0.00 \, \text{V} = 0.80 \, \text{V} \]

### Step 4: Calculate the Gibbs free energy change
The Gibbs free energy change (ΔG) can be calculated using the relationship:
\[ \Delta G = -nFE^\circ_{cell} \]
Where:
- n = number of moles of electrons transferred (1 mole in this case)
- F = Faraday's constant (approximately 96485 C/mol)

Thus:
\[ \Delta G = -1 \times 96485 \, \text{C/mol} \times 0.80 \, \text{V} \]
\[ \Delta G = -77188 \, \text{J/mol} \quad \text{or} \quad -77.2 \, \text{kJ/mol} \]

### Step 5: Conclusion
The overall reaction in the galvanic cell has a standard EMF of 0.80 V, and the Gibbs free energy change is -77.2 kJ/mol, indicating that the reaction is spontaneous.

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