Assume a cell with the following reaction Cu(s) + 2Ag+(1 x 10^-3 M) - Cu^2+(0.250 M) + 2Ag(s), E^o_

Q: Why don't we include solid Copper and Silver in the reaction quotient $Q$?

Pure solids and liquids have a constant concentration (or activity of 1) throughout the reaction. Including them wouldn't change the ratio, so they are omitted from the $Q$ expression.

Q: Why is the concentration of $Ag^+$ squared in the denominator?

In the balanced chemical equation, the stoichiometric coefficient of $Ag^+$ is 2. According to the law of mass action, concentrations in the reaction quotient must be raised to the power of their coefficients.

Q: How did we know that $n = 2$?

Look at the oxidation states. Copper goes from 0 to +2, losing 2 electrons. Silver goes from +1 to 0, but there are two Silver atoms reacting. Thus, a total of 2 electrons are transferred between the species.

Q: What does the standard cell potential $E_{cell}^{\circ}$ actually mean?

It is the maximum voltage the cell can produce when all reactants and products are at standard conditions, which means 1 Molar concentration for solutions and 1 atm pressure for gases at 298 K.

Q: Why does the cell potential decrease over time?

As the reaction proceeds forward, reactants ($Ag^+$) are consumed and products ($Cu^{2+}$) are formed. This increases the value of $Q$. Since we subtract the log of $Q$ in the Nernst equation, a larger $Q$ results in a smaller $E_{cell}$.

Question

Assume a cell with the following reaction
$Cu(s) + 2Ag^+(1 	imes 10^{-3} M) \longrightarrow Cu^{2+}(0.250 M) + 2Ag(s)$,
$E_{cell}^{\circ} = 2.97 V$
$E_{cell}$ for the above reaction is ......... V.
(Nearest integer)
[Given : $\log 2.5 = 0.3979, T = 298 K$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a microscopic battle for electrons. On one side, we have a solid Copper electrode, and on the other, a Silver electrode.

The given reaction is:
$$Cu(s) + 2Ag^+(aq) \longrightarrow Cu^{2+}(aq) + 2Ag(s)$$

Here, Copper is acting as the **anode**. It is willingly giving up its electrons, undergoing oxidation to become $Cu^{2+}$ ions.

Meanwhile, the Silver ions in the solution are hungry for those electrons. They grab them and deposit as solid Silver on the **cathode**.

This flow of electrons from Copper to Silver is what generates the electrical potential, or voltage, of our cell.

### The Master Equation

We are given the standard cell potential, $E_{cell}^{\circ} = 2.97 V$. But our cell is not at standard conditions! The concentrations of the ions are different.

To find the actual voltage, we must invoke the **Nernst Equation**:
$$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q$$

This beautiful equation adjusts the standard potential based on the actual ratio of products to reactants, known as the reaction quotient, $Q$.

### Finding the Missing Pieces

Before we plug in the numbers, we need to determine two critical values: $n$ and $Q$.

The value of $n$ represents the total number of moles of electrons transferred in the balanced reaction. Since Copper loses 2 electrons and two Silver ions gain 1 electron each (totaling 2), we have **$n = 2$**.

Next, we write the expression for $Q$. Remember, pure solids like $Cu(s)$ and $Ag(s)$ do not appear in the quotient. 
$$Q = \frac{[Cu^{2+}]}{[Ag^+]^2}$$

**Crucial Trap:** Notice the squared term in the denominator! Because the stoichiometric coefficient of $Ag^+$ is 2, its concentration must be squared. Forgetting this is a classic silly mistake.

### The Final Calculation

Now, let's substitute our known values into the Nernst Equation:
$$E_{cell} = 2.97 - \frac{0.0591}{2} \log \frac{0.250}{(10^{-3})^2}$$

Let's simplify the logarithmic term carefully. The denominator becomes $10^{-6}$. Bringing it to the numerator gives us:
$$\log (0.250 	imes 10^6) = \log (2.5 	imes 10^5)$$

Using the properties of logarithms, we can split this:
$$\log (2.5 	imes 10^5) = \log 2.5 + \log 10^5 = 0.3979 + 5 = 5.3979$$

Finally, we plug this back into our main equation:
$$E_{cell} = 2.97 - \frac{0.0591}{2} 	imes 5.3979$$
$$E_{cell} = 2.97 - 0.1595 = 2.8105 V$$

The question asks for the nearest integer. Rounding $2.8105 V$ gives us our final answer: **$3 V$**.

As this cell continues to run, the concentration of Copper ions will increase, and Silver ions will decrease, causing the voltage to slowly drop until it reaches zero at equilibrium.

Enter Numerical Value:

Visualized Solution

$C u (s) + 2 A g^{+} (a q) ⟶ C u^{2 +} (a q) + 2 A g (s)$

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

$F in d in g n an d Q$

$S u b s t i t u t i o n$

$S im pl i f y in g t h e L o g T er m$

$C a l c u l a t in g P o t e n t ia l$

$F ina l A n s w er$

$D y nami c E q u i l ib r i u m$

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

Finding the Missing Pieces

The Final Calculation

Similar Questions

Given the equilibrium constant ( $K_{c}$ ) of the reaction : $C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)$ is $10 \times 1 0^{15}$ , calculate the $E_{ce l l}^{\circ}$ of this reaction at 298 K. $[2.303 \frac{R T}{F} at 298 K = 0.059 V]$

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

For the galvanic cell, $Z n (s) + C u^{2 +} (0.02 M) ⟶ Z n^{2 +} (0.04 M) + C u (s)$ , $E_{C e l l} = \dots \times 1 0^{- 2} V$ . (Nearest integer) [Use $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction at $2 5^{\circ} C$ will be

The standard Gibbs energy for the given cell reaction in $kJ mol^{- 1}$ at $298 K$ is $Zn (s) + Cu^{2 +} (a q) ⟶ Zn^{2 +} (a q) + Cu (s), E^{\circ} = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C mol^{- 1}$ )

For a cell given below $A g ∣ A g^{+} ∣∣ C u^{2 +} ∣ C u$ $A g^{+} + e^{-} ⟶ A g; E^{\circ} = x$ $C u^{2 +} + 2 e^{-} ⟶ C u; E^{\circ} = y$ $E^{\circ}$ cell is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cu(s)+2Ag+(aq)⟶Cu2+(aq)+2Ag(s)

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

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

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

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

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

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

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