What is the sign of `DeltaG^@` and the value of K an electrochemical cell for which `E_("cell")^@=0.80V` ?

Options

A. `DeltaG=-ve`
`kgt1`
B. `DeltaG=+ve`
`kgt1`
C. `DeltaG=+ve`
`klt1`
D. `DeltaG=-ve`
`klt1`

Answer

To solve the problem, we need to determine the sign of ΔG° and the value of K for an electrochemical cell with a standard cell potential (E°cell) of 0.80 V.

### Step-by-Step Solution:

1. **Understanding the Relationship Between ΔG° and E°cell**:
The Gibbs free energy change (ΔG°) for an electrochemical reaction is related to the standard cell potential (E°cell) by the equation:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) = number of moles of electrons transferred in the reaction,
- \( F \) = Faraday's constant (approximately 96500 C/mol),
- \( E^\circ_{\text{cell}} \) = standard cell potential.

2. **Substituting the Given Value**:
Since we are given that \( E^\circ_{\text{cell}} = 0.80 \, \text{V} \), we can analyze the sign of ΔG°:
- The value of \( E^\circ_{\text{cell}} \) is positive (0.80 V).
- Therefore, since \( n \) and \( F \) are both positive constants, the product \( nFE^\circ_{\text{cell}} \) is also positive.

3. **Determining the Sign of ΔG°**:
Substituting into the equation:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} < 0 \]
This indicates that ΔG° is negative.

4. **Relating ΔG° to the Equilibrium Constant K**:
The relationship between ΔG° and the equilibrium constant (K) is given by:
\[ \Delta G^\circ = -RT \ln K \]
where:
- \( R \) = universal gas constant (8.314 J/(mol·K)),
- \( T \) = temperature in Kelvin.

5. **Analyzing the Sign of K**:
Since we have established that ΔG° is negative, we can substitute this into the equation:
\[ -RT \ln K < 0 \]
This implies that \( \ln K > 0 \), which means:
\[ K > 1 \]

### Final Results:
- The sign of ΔG° is **negative**.
- The value of K is **greater than 1** (K > 1).