Calculate the EMF of the following cell: $Cd \mid Cd^{+2}(1\,M) \parallel Ag^{+}(1\,M) \mid Ag$, for which the standard reduction potentials of the half-reactions are: $Cd^{+2}\mid Cd;\ E^{\circ} = -0.40\,V$ and $Ag^{+}\mid Ag;\ E^{\circ} = +0.80\,V$.

Question

Accepted Answer

>

> [!TLDR]
> $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.80-(-0.40)=1.20\,V$ (all species $1\,M$, so $E=E^\circ$).
>
> **Concept:** In the cell $Cd\mid Cd^{2+}\parallel Ag^+\mid Ag$, the electrode written on the left is the **anode** (oxidation, $Cd\rightarrow Cd^{2+}+2e^-$) and the right is the **cathode** (reduction, $Ag^++e^-\rightarrow Ag$). The cell EMF is
> $$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}.$$
>
> **Substitute** the standard reduction potentials: …

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

Q.Calculate the EMF of the following cell: $Cd \mid Cd^{+2}(1\,M) \parallel Ag^{+}(1\,M) \mid Ag$ , for which the standard reduction potentials of the half-reactions are: $Cd^{+2}\mid Cd;\ E^{\circ} = -0.40\,V$ and $Ag^{+}\mid Ag;\ E^{\circ} = +0.80\,V$ .

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

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