Nernst Equation 🚀
Picture a battery where tiny ions race around, changing their concentration and, in turn, the volt-meter reading. The Nernst equation tells you exactly how that voltage changes. 🔋⚡ :contentReference[oaicite:0]{index=0}
1 · Starting Point: A Simple Half-Reaction 🧪
Consider \( \mathrm{M}^{n+}(aq) + n e^- \;\longrightarrow\; M(s) \) Then $$ E = E^{\circ} – \frac{RT}{nF}\,\ln\!\bigl(\tfrac{1}{[\mathrm{M}^{n+}]}\bigr) $$ Here \(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\), \(T\) is kelvin, \(F = 96487\ \text{C mol}^{-1}\), and \(n\) equals the electrons moved. 💡 :contentReference[oaicite:1]{index=1}
2 · Handy Shortcut at 298 K 😎
At room temperature \(T = 298\ \text{K}\),
\( \dfrac{RT}{F} = 0.059\ \text{V} \).
So the half-cell equation becomes
$$
E = E^{\circ} – \frac{0.059}{n}\,\log\!\bigl(\tfrac{1}{[\mathrm{M}^{n+}]}\bigr)
$$
Just remember “0.059 / n” and you’re ready for quick sums in seconds! ⏱️ :contentReference[oaicite:2]{index=2}
3 · Putting Two Electrodes Together: Daniell Cell Example
Cathode (\(\mathrm{Cu}^{2+}/\mathrm{Cu}\)) potential: \( E_{\mathrm{Cu}} = E^{\circ}_{\mathrm{Cu}} – \dfrac{RT}{2F}\ln\bigl(\tfrac{1}{[\mathrm{Cu}^{2+}]}\bigr) \) :contentReference[oaicite:3]{index=3}
Anode (\(\mathrm{Zn}^{2+}/\mathrm{Zn}\)) potential: \( E_{\mathrm{Zn}} = E^{\circ}_{\mathrm{Zn}} – \dfrac{RT}{2F}\ln\bigl(\tfrac{1}{[\mathrm{Zn}^{2+}]}\bigr) \) :contentReference[oaicite:4]{index=4}
Cell voltage: $$ E_{\text{cell}} \;=\; E^{\circ}_{\text{cell}} – \frac{RT}{2F}\,\ln\!\frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} $$ At 298 K this simplifies to \( E_{\text{cell}} = E^{\circ}_{\text{cell}} – \dfrac{0.059}{2}\,\log\!\bigl(\tfrac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}\bigr) \). 🧮 :contentReference[oaicite:5]{index=5}
4 · General Form for ANY Cell
For a reaction \( aA + bB + ne^- \;\rightleftharpoons\; cC + dD \) the reaction quotient \(Q\) equals \( Q = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \). Nernst gives $$ E_{\text{cell}} = E^{\circ}_{\text{cell}} – \frac{RT}{nF}\,\ln Q $$ or at 298 K \( E_{\text{cell}} = E^{\circ}_{\text{cell}} – \dfrac{0.059}{n}\,\log Q \). 📐 :contentReference[oaicite:6]{index=6}
5 · Linking Voltage to Equilibrium 🔄
When a galvanic cell relaxes to equilibrium, \(E_{\text{cell}} = 0\) and \(Q = K_c\). Set those into the Nernst expression to get $$ E^{\circ}_{\text{cell}} = \frac{2.303\,RT}{nF}\,\log K_c $$ At 298 K that’s \( E^{\circ}_{\text{cell}} = \dfrac{0.059}{n}\,\log K_c \). ✨ :contentReference[oaicite:7]{index=7}
6 · Voltage and Free-Energy 💚
A running cell does work equal to the drop in Gibbs energy: \( \Delta_r G = -\,n F E_{\text{cell}} \). Under standard conditions: \( \Delta_r G^{\circ} = -\,n F E^{\circ}_{\text{cell}} \). 🔋 :contentReference[oaicite:8]{index=8}
7 · What Standard Potentials Tell You
- Positive \(E^{\circ}\): the reduced side beats \(\mathrm{H}_2\) as a winner of electrons — strong oxidising agent. 🤝 :contentReference[oaicite:9]{index=9}
- Negative \(E^{\circ}\): the metal loves to give up electrons — strong reducing agent. 💪 :contentReference[oaicite:10]{index=10}
- \(\mathrm{F_2}\) sits highest (+2.87 V), making fluoride the weakest reducer and fluorine the fiercest oxidiser; \(\mathrm{Li}\) lies lowest (–3.05 V), so lithium metal is your champion reducer. 🏆 :contentReference[oaicite:11]{index=11}
- The table of values helps you pick the right metal for redox, calculate pH, run potentiometric titrations, and more. 🧑🔬 :contentReference[oaicite:12]{index=12}
8 · Quick Worked Example 📝
Cell: \(\mathrm{Mg}(s) \,|\, \mathrm{Mg}^{2+}(0.130\,\text{M}) \,\|\, \mathrm{Ag}^{+}(0.0001\,\text{M}) \,|\, \mathrm{Ag}(s)\) Given \(E^{\circ}_{\text{cell}} = 3.17\,\text{V}\) and \(n = 2\): $$ E_{\text{cell}} = 3.17\ \text{V} \;-\; \frac{0.059}{2}\,\log\!\Bigl(\tfrac{0.130}{(0.0001)^2}\Bigr) = 2.96\ \text{V} $$ so the magnesium-silver combo packs quite the punch! ⚡ :contentReference[oaicite:13]{index=13}
Important Concepts for NEET 🔥
- The magic shortcut \(0.059/n\) makes Nernst calculations lightning-fast at 298 K. 💡 :contentReference[oaicite:14]{index=14}
- Free-energy link: \( \Delta_r G^{\circ} = -n F E^{\circ}_{\text{cell}} \). 🔋 :contentReference[oaicite:15]{index=15}
- Equilibrium constant relation: \( E^{\circ}_{\text{cell}} = (0.059/n)\,\log K_c \). 📈 :contentReference[oaicite:16]{index=16}
- Sign of \(E^{\circ}\) predicts oxidising vs. reducing strength. ⚖️ :contentReference[oaicite:17]{index=17}
- Remember to balance electrons (n) for both electrodes before using Nernst. 🧮 :contentReference[oaicite:18]{index=18}
Keep practicing — every log brings you closer to full marks! 🎯
