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 given electrochemical cell:

**Step 1: Identify the half-reactions.**
- The oxidation half-reaction involves zinc:
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
- The reduction half-reaction involves silver:
\[ \text{Ag}^{+}(aq) + e^- \rightarrow \text{Ag(s)} \]

**Step 2: Write the balanced net cell reaction.**
- Since two electrons are produced in the oxidation of zinc, we need to multiply the reduction half-reaction by 2 to balance the electrons:
\[ 2\text{Ag}^{+}(aq) + 2e^- \rightarrow 2\text{Ag(s)} \]
- Combining both half-reactions gives us the overall cell reaction:
\[ \text{Zn(s)} + 2\text{Ag}^{+}(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag(s)} \]

**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 coefficients in the balanced equation:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} \]

**Step 4: Substitute the values into the expression.**
- From the problem, we have:
- \([\text{Zn}^{2+}] = 0.01 \, M\)
- \([\text{Ag}^{+}] = 1.25 \, M\)
- Substituting these values into the expression for \( Q \):
\[ Q = \frac{0.01}{(1.25)^2} \]

**Step 5: Calculate \( Q \).**
- First, calculate \( (1.25)^2 \):
\[ (1.25)^2 = 1.5625 \]
- Now substitute this back into the equation for \( Q \):
\[ Q = \frac{0.01}{1.5625} \approx 0.0064 \]
- To express this in scientific notation:
\[ Q \approx 64 \times 10^{-3} \]

**Final Answer:**
The value of the reaction quotient \( Q \) for the cell is approximately \( 64 \times 10^{-3} \).

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