For the cell Cu(s)|Cu2+(aq)(0.1 M)|| Ag+(aq)(0.01M)| Ag(s) the cell potential, E1 = 0.3095 V. For th

Q: Why is $n=2$ in this reaction?

Because copper oxidizes from 0 to +2 (losing 2 electrons), and two silver ions reduce from +1 to 0 (gaining 1 electron each, total 2).

Q: Why do we square the concentration of $Ag^+$ in the Nernst equation?

The stoichiometric coefficient of $Ag^+$ in the balanced cell reaction is 2. In the reaction quotient $Q$, concentrations are raised to the power of their stoichiometric coefficients.

Q: Can we calculate $E^\circ_{cell}$ as a decimal first?

Yes, but keeping it as an expression saves time and reduces rounding errors in intermediate steps.

Q: What happens if the cell potential becomes negative?

A negative cell potential means the reaction is non-spontaneous in the forward direction, and the cell would act as an electrolytic cell requiring external voltage.

Q: Why is the concentration of solid Cu and Ag not included in $Q$?

The concentration (or activity) of pure solids and liquids is taken as unity (1) and does not change during the reaction.

Question

For the cell
$Cu(s)|Cu^{2+}(aq)(0.1 M)|| Ag^+(aq)(0.01M)| Ag(s)$
the cell potential, $E_1 = 0.3095 V$
For the cell,
$Cu(s)|Cu^{2+}(aq)(0.01 M)|| Ag^+(aq)(0.001M)| Ag(s)$
the cell potential
$= \dots \dots \dots 	imes 10^{-2} V$.
(Round off the nearest integer).
$\left[ 	ext{Use: } \frac{2.303 RT}{F} = 0.059 ight]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a galvanic cell where copper acts as the anode and silver acts as the cathode. At the anode, copper undergoes oxidation, losing two electrons to form $Cu^{2+}$ ions. Meanwhile, at the cathode, two $Ag^+$ ions each gain an electron to form solid silver.

The overall cell reaction is:
$$Cu(s) + 2Ag^+(aq) ightarrow Cu^{2+}(aq) + 2Ag(s)$$

From this balanced equation, we can clearly see that the number of electrons transferred during the process is $n = 2$. This is a crucial piece of information for our next step.

### The Master Equation

To find the cell potential at any given concentration, we rely on the Nernst equation. It connects the standard cell potential to the actual cell potential based on the reaction quotient $Q$.

$$E_{cell} = E^\circ_{cell} - \frac{2.303 RT}{nF} \log Q$$

Given in the problem, we can substitute $\frac{2.303 RT}{F} = 0.059$. The reaction quotient $Q$ is the ratio of the concentration of the products to the reactants, raised to their stoichiometric coefficients. Since solids are not included, $Q = \frac{[Cu^{2+}]}{[Ag^+]^2}$.

$$E_{cell} = E^\circ_{cell} - \frac{0.059}{2} \log \frac{[Cu^{2+}]}{[Ag^+]^2}$$

### Evaluating the First Cell

For the first cell, we are given the concentrations: $[Cu^{2+}] = 0.1 	ext{ M}$ and $[Ag^+] = 0.01 	ext{ M}$. The measured cell potential is $E_1 = 0.3095 	ext{ V}$. Let's substitute these values into our Nernst equation.

$$0.3095 = E^\circ_{cell} - \frac{0.059}{2} \log \frac{0.1}{(0.01)^2}$$

Now, let's simplify the logarithmic term. The numerator is $10^{-1}$ and the denominator is $(10^{-2})^2 = 10^{-4}$.

$$0.3095 = E^\circ_{cell} - \frac{0.059}{2} \log \frac{10^{-1}}{10^{-4}}$$
$$0.3095 = E^\circ_{cell} - \frac{0.059}{2} \log(10^3)$$

Using the property of logarithms, $\log(10^3) = 3$.

$$E^\circ_{cell} = 0.3095 + \frac{0.059 	imes 3}{2}$$

**Pro Tip:** Do not evaluate this expression just yet! Keeping it in this raw form will make our final calculation much cleaner and less prone to rounding errors.

### Setting Up the Second Cell

Now, let's look at the second cell where the concentrations have been altered. The new concentrations are $[Cu^{2+}] = 0.01 	ext{ M}$ and $[Ag^+] = 0.001 	ext{ M}$. We need to find the new cell potential, $E_2$.

$$E_2 = E^\circ_{cell} - \frac{0.059}{2} \log \frac{0.01}{(0.001)^2}$$

Again, we simplify the logarithmic term carefully. The numerator is $10^{-2}$ and the denominator is $(10^{-3})^2 = 10^{-6}$.

$$E_2 = E^\circ_{cell} - \frac{0.059}{2} \log \frac{10^{-2}}{10^{-6}}$$
$$E_2 = E^\circ_{cell} - \frac{0.059}{2} \log(10^4)$$

This simplifies to:

$$E_2 = E^\circ_{cell} - \frac{0.059 	imes 4}{2}$$

### Final Calculation

This is where the magic happens. We substitute our raw expression for $E^\circ_{cell}$ from the first cell directly into the equation for the second cell.

$$E_2 = \left( 0.3095 + \frac{0.059 	imes 3}{2} ight) - \frac{0.059 	imes 4}{2}$$

Notice how beautifully the terms combine! We can factor out the common $\frac{0.059}{2}$.

$$E_2 = 0.3095 + \frac{0.059}{2} (3 - 4)$$
$$E_2 = 0.3095 - \frac{0.059}{2}$$

Now, we just perform the final arithmetic.

$$E_2 = 0.3095 - 0.0295 = 0.2800 	ext{ V}$$

The question asks for the answer in the format $\dots 	imes 10^{-2} 	ext{ V}$.

$$0.2800 	ext{ V} = \mathbf{28 	imes 10^{-2} 	ext{ V}}$$

Thus, the final integer answer is **28**.

Enter Numerical Value:

Visualized Solution

$C e l l R e a c t i o n$

$N er n s tE q u a t i o n$

$S e t u p f or C e l l 1$

$C o m p u t in g E_{ce l l}^{\circ}$

$S e t u p f or C e l l 2$

$C o m p u t in g C e l l 2 P o t e n t ia l$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

Evaluating the First Cell

Setting Up the Second Cell

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CellReaction

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheWayForward