The value of reaction quotient `Q` for the cell
`Zn(s)|Zn^(2+)(0.01M)||Ag^(o+)(1.25M)|Ag(s) is `

Options

A. 156
B. 125
C. 1.25xx10^(-2)
D. 64xx10^(-3)

Explanation

To find the value of the reaction quotient \( Q \) for the cell represented as \( \text{Zn(s)}|\text{Zn}^{2+}(0.01M)||\text{Ag}^{+}(1.25M)|\text{Ag(s)} \), we will follow these steps:

### Step 1: Identify the half-reactions
In the given electrochemical cell:
- The anode reaction (oxidation) involves zinc:
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+} + 2e^- \]
- The cathode reaction (reduction) involves silver:
\[ \text{Ag}^{+} + e^- \rightarrow \text{Ag(s)} \]
To balance the electrons, we need to multiply the silver half-reaction by 2:
\[ 2\text{Ag}^{+} + 2e^- \rightarrow 2\text{Ag(s)} \]

### Step 2: Write the overall cell reaction
Combining the half-reactions gives us the overall cell reaction:
\[ \text{Zn(s)} + 2\text{Ag}^{+} \rightarrow \text{Zn}^{2+} + 2\text{Ag(s)} \]

### Step 3: Write the expression for the reaction quotient \( Q \)
The reaction quotient \( Q \) is defined as the ratio of the concentrations of the products to the concentrations of the reactants, raised to the power of their stoichiometric coefficients:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} \]

### Step 4: Substitute the given concentrations into the expression
From the problem, we know:
- The concentration of \( \text{Zn}^{2+} = 0.01 \, M \)
- The concentration of \( \text{Ag}^{+} = 1.25 \, M \)

Substituting these values into the expression for \( Q \):
\[ Q = \frac{0.01}{(1.25)^2} \]

### Step 5: Calculate \( Q \)
Now, calculate \( (1.25)^2 \):
\[ (1.25)^2 = 1.5625 \]
Then substitute this back into the expression for \( Q \):
\[ Q = \frac{0.01}{1.5625} \approx 0.0064 \]

### Step 6: Express \( Q \) in scientific notation
To express \( 0.0064 \) in scientific notation:
\[ Q = 6.4 \times 10^{-3} \]

### Final Answer
Thus, the value of the reaction quotient \( Q \) for the cell is:
\[ Q \approx 6.4 \times 10^{-3} \]