For a cell reaction involving a two electron change, the standrard emf of the cell is found to be `0.295V" at "25^(@)`C. The equilibrium constant of the reaction at `25^(@)`C will be:

Options

A. `1 xx 10^(-10)`
B. `29.5xx 10^(-2)`
C. 10
D. `1xx10^(10)`

Answer

To find the equilibrium constant (Kc) of a cell reaction involving a two-electron change with a standard emf (E°) of 0.295 V at 25°C, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the Nernst Equation
The Nernst equation relates the standard cell potential (E°) to the cell potential (E) and the equilibrium constant (Kc):
\[ E = E° - \frac{0.0591}{n} \log Kc \]
where:
- E = cell potential at equilibrium (0 V)
- E° = standard cell potential (0.295 V)
- n = number of electrons transferred in the reaction (2 in this case)

### Step 2: Set E to 0 at Equilibrium
At equilibrium, the cell potential (E) is 0 V. Thus, we can rewrite the equation:
\[ 0 = 0.295 - \frac{0.0591}{2} \log Kc \]

### Step 3: Rearrange the Equation
Rearranging the equation to solve for log Kc gives:
\[ \frac{0.0591}{2} \log Kc = 0.295 \]
\[ \log Kc = \frac{0.295 \times 2}{0.0591} \]

### Step 4: Calculate log Kc
Now, calculate the right-hand side:
\[ \log Kc = \frac{0.59}{0.0591} \]
Calculating this gives:
\[ \log Kc \approx 10.0 \]

### Step 5: Convert log Kc to Kc
To find Kc, we use the property of logarithms:
\[ Kc = 10^{\log Kc} = 10^{10} \]

### Final Answer
Thus, the equilibrium constant (Kc) of the reaction at 25°C is:
\[ Kc = 1 \times 10^{10} \]