For a cell given below
Ag | Ag+ || Cu2+ | Cu
Ag+ + e- - Ag; E° = x
Cu2+ + 2e- - Cu; E° = y
E° cell

Q: Why don't we multiply the potential $x$ by 2 since we multiplied the silver reaction by 2?

This is a classic trap! Standard electrode potential ($E^\circ$) is an intensive property, meaning it depends on the nature of the substance, not the quantity. Just like temperature doesn't double if you have twice as much water, the potential doesn't double if you have twice as many atoms reacting.

Q: How do I know which electrode is the anode and which is the cathode just from the notation?

Always remember the ABC rule for cell notation: **A**node | **B**ridge | **C**athode. The species written on the left of the double vertical lines (salt bridge) is always the anode, and the one on the right is the cathode.

Q: The formula for cell potential is $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$. Why did we add them in the solution?

Both methods are perfectly valid! The formula $E^\circ_{cathode} - E^\circ_{anode}$ uses standard **reduction** potentials for both. So, $y - x$. In our step-by-step solution, we converted the anode's potential to an **oxidation** potential ($-x$) and then simply added it to the cathode's reduction potential ($y$). The result is exactly the same: $y - x$.

Q: Wait, isn't silver less reactive than copper? Shouldn't copper be the anode?

You are absolutely right about their chemical reactivity! In a real, spontaneous galvanic cell, copper would be the anode and silver the cathode. However, in this specific problem, we are asked to calculate the potential for the cell *exactly as it is written*. The resulting negative potential simply tells us that the cell reaction is non-spontaneous in the forward direction.

Q: What does the single vertical line $|$ represent in the cell notation?

The single vertical line represents a phase boundary. It separates the solid electrode (like $Ag_{(s)}$) from its aqueous ions (like $Ag^+_{(aq)}$) in the solution.

Question

For a cell given below
$Ag | Ag^+ || Cu^{2+} | Cu$
$Ag^+ + e^- \longrightarrow Ag; E^\circ = x$
$Cu^{2+} + 2e^- \longrightarrow Cu; E^\circ = y$
$E^\circ$ cell is

Options:

(A) $x + 2y$

(B) $2x + y$

(C) $y - x$

(D) $y - 2x$

Sigma Percentile · Accepted Answer

Electrochemical cells are the beating heart of modern portable energy, and understanding how to calculate their potential is a fundamental skill in chemistry. This problem is a beautiful exercise in decoding cell notation and avoiding one of the most common traps in electrochemistry: the intensive nature of standard electrode potentials.

Let's embark on a journey to break down this cell and calculate its standard potential step-by-step.

### Decoding the Cell Notation

The problem presents us with a specific cell notation:

$$Ag | Ag^+ || Cu^{2+} | Cu$$

To understand what is happening, we must rely on the universal **ABC convention** of cell notation: **A**node | **B**ridge | **C**athode.

This notation explicitly tells us that the Silver ($Ag$) electrode is acting as the **anode** (placed on the left), and the Copper ($Cu$) electrode is acting as the **cathode** (placed on the right). The double vertical lines ($||$) represent the salt bridge connecting the two half-cells.

### Analyzing the Half-Reactions

Now that we know who is doing what, let's look at the chemistry happening at each electrode.

**At the Anode (Oxidation):**
By definition, oxidation (the loss of electrons) occurs at the anode. Therefore, solid silver is losing an electron to become a silver ion:

$$Ag \longrightarrow Ag^+ + e^-$$

The problem provides the standard *reduction* potential for silver as $x$. However, since silver is undergoing *oxidation* here, we must reverse the sign of the potential. Thus, the standard oxidation potential is:

$$E^\circ_{oxi} = -x$$

**At the Cathode (Reduction):**
Reduction (the gain of electrons) occurs at the cathode. Copper ions in the solution are gaining two electrons to deposit as solid copper:

$$Cu^{2+} + 2e^- \longrightarrow Cu$$

The problem directly gives us the standard reduction potential for this exact process as $y$. We don't need to change anything here:

$$E^\circ_{red} = y$$

### The Trap of Intensive Properties

To find the overall cell reaction, we must ensure that the number of electrons lost equals the number of electrons gained. Copper requires two electrons, but silver only provides one. To balance this, we must multiply the silver half-reaction by 2:

$$2Ag \longrightarrow 2Ag^+ + 2e^-$$

Here lies the greatest trap in electrochemistry! You might be tempted to multiply the oxidation potential ($-x$) by 2 as well. **Do not do this.**

Standard electrode potential ($E^\circ$) is an **intensive property**. It depends solely on the nature of the chemical species involved and their standard states, not on the stoichiometric quantities reacting. Just as the boiling point of water doesn't double if you boil two liters instead of one, the electrical pressure (potential) doesn't double if two moles of silver react instead of one.

Therefore, even after balancing, the oxidation potential remains strictly:

$$E^\circ_{oxi} = -x$$

### Calculating the Final Potential

The standard cell potential ($E^\circ_{cell}$) is simply the sum of the oxidation potential at the anode and the reduction potential at the cathode:

$$E^\circ_{cell} = E^\circ_{oxi} + E^\circ_{red}$$

Substituting our values:

$$E^\circ_{cell} = -x + y$$

Rearranging this gives us our final, elegant answer:

$$E^\circ_{cell} = y - x$$

*A fascinating reality check:* If you look up the actual standard reduction potentials, silver ($+0.80	ext{ V}$) is higher than copper ($+0.34	ext{ V}$). In a real-world spontaneous cell, copper would be the anode and silver the cathode. The cell as written in the problem would actually have a negative potential, meaning it is non-spontaneous. However, mathematically and strictly following the provided notation, the expression $y - x$ is flawlessly correct.

Animated Solution for Chemistry - Electrochemistry: 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

Select Answer:

Visualized Solution

Decoding the Cell Notation

Anode Reaction (Oxidation)

Cathode Reaction (Reduction)

Balancing Electrons

Standard Cell Potential

The Sigma Insight: Electrochemical Cells

Decoding the Cell Notation

Analyzing the Half-Reactions

The Trap of Intensive Properties

Calculating the Final Potential

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For the given cell; $C u (s) ∣ C u^{2 +} (C_{1} M) ∣∣ C u^{2 +} (C_{2} M) ∣ C u (s)$ change in Gibbs energy ( $Δ G$ ) is negative, if

The cell, $Zn ∣ Zn^{2 +} (1 M) ∣∣ Cu^{2 +} (1 M) ∣ Cu$ ( $E_{cell}^{\circ} = 1.10 V$ ), was allowed to be completely discharged at 298 K. The relative concentration of $Zn^{2 +}$ to $Cu^{2 +}$ $(\frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]})$ is

Animated Solution for Chemistry - Electrochemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a cell given below Ag∣Ag+∣∣Cu2+∣Cu Ag++e−⟶Ag;E∘=x Cu2++2e−⟶Cu;E∘=y E∘ cell is

Select Answer:

Visualized Solution

Decoding the Cell Notation

Anode Reaction (Oxidation)

Cathode Reaction (Reduction)

Balancing Electrons

Standard Cell Potential

The Sigma Insight: Electrochemical Cells

Decoding the Cell Notation

Analyzing the Half-Reactions

The Trap of Intensive Properties

Calculating the Final Potential

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Animated Solution for Chemistry - Electrochemistry: 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

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