Consider the cell reaction `Zn+Cu^(2+)(aq)iff Cu + Zn^(2+)` (aq). Reaction quotient is `Q = ([Zn^(2+)])/([Cu^(2+)])E^(0)` of the cell is `1.10V.` Now, `E_(cell)` will be `1.159V` when:

Options

A. `([Cu^(2+)])/([Zn^(2+)]) = 0.01`
B. `([Zn^(2+)])/([Cu^(2+)]) = 0.01`
C. `([Zn^(2+)])/([Cu^(2+)]) = 0.1`
D. `([Zn^(2+)])/([Cu^(2+)]) = 1`

Answer

To solve the problem step by step, we will use the Nernst equation and the information provided in the question.

### Step 1: Write down the Nernst equation.
The Nernst equation relates the cell potential (E_cell) to the standard cell potential (E°_cell), the number of moles of electrons transferred (n), and the reaction quotient (Q).

\[ E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q \]

### Step 2: Identify the values given in the question.
From the question, we have:
- E°_cell = 1.10 V
- E_cell = 1.159 V
- The reaction involves the transfer of 2 electrons (n = 2) because Zn is oxidized to Zn²⁺ and Cu²⁺ is reduced to Cu.

### Step 3: Substitute the known values into the Nernst equation.
Substituting the known values into the Nernst equation:

\[ 1.159 = 1.10 - \frac{0.0591}{2} \log Q \]

### Step 4: Rearrange the equation to solve for log Q.
First, isolate the log Q term:

\[ 1.159 - 1.10 = - \frac{0.0591}{2} \log Q \]

This simplifies to:

\[ 0.059 = - \frac{0.0591}{2} \log Q \]

### Step 5: Multiply both sides by -2 to eliminate the fraction.
\[ -2 \times 0.059 = 0.0591 \log Q \]

This gives:

\[ -0.118 = 0.0591 \log Q \]

### Step 6: Solve for log Q.
Now, divide both sides by 0.0591:

\[ \log Q = \frac{-0.118}{0.0591} \]

Calculating this gives:

\[ \log Q \approx -2 \]

### Step 7: Convert from log Q to Q.
To find Q, we use the property of logarithms:

\[ Q = 10^{-2} = 0.01 \]

### Step 8: Write the expression for Q.
The reaction quotient Q is given by:

\[ Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \]

From the calculated value of Q:

\[ \frac{[Zn^{2+}]}{[Cu^{2+}]} = 0.01 \]

### Step 9: Interpret the result.
This means that the concentration of Zn²⁺ is 0.01 times the concentration of Cu²⁺. If we let [Cu²⁺] = x, then [Zn²⁺] = 0.01x.

### Final Answer:
The concentrations must satisfy the ratio derived from Q, which indicates that the concentration of Zn²⁺ is much lower than that of Cu²⁺.

---