Masterclass: 25 Solved JEE Advanced Numericals on the Nernst Equation
The ultimate problem-solving guide covering concentration cells, thermodynamics, pH gradients, solubility products, and complexation equilibria.
Welcome to the most rigorous and exhaustive collection of solved numerical problems on the Nernst Equation available online, exclusively on Chemca.in. If you are preparing for JEE Advanced or NEET, you know that simply memorizing the formula is entirely insufficient. These competitive exams test your ability to link cell potential with chemical equilibrium, ionic gradients, and thermodynamics.
Before attempting these high-level problems, we strongly recommend mastering the foundational calculus and thermodynamic origins of the formula. Ensure your basics are rock solid by studying our Exhaustive Guide: Derivation of the Nernst Equation.
In this guide, we will tackle 25 meticulously designed problems. We assume standard temperature ($298 \text{ K}$) unless otherwise specified, allowing us to use the simplified constant $\frac{0.0591}{n}$ in our calculations.
Question: Calculate the EMF of the following cell at $298 \text{ K}$:
$Zn_{(s)} | Zn^{2+} (0.01 \text{ M}) || Ag^+ (0.1 \text{ M}) | Ag_{(s)}$
Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \text{ V}$ and $E^{\circ}_{Ag^+/Ag} = +0.80 \text{ V}$.
Anode: $Zn \rightarrow Zn^{2+} + 2e^-$
Cathode: $2Ag^+ + 2e^- \rightarrow 2Ag$
Overall: $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$
Number of electrons transferred, $n = 2$.
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.80 - (-0.76) = 1.56 \text{ V}$
$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log_{10} \frac{[Zn^{2+}]}{[Ag^+]^2}$
$E_{\text{cell}} = 1.56 - 0.02955 \log_{10} \frac{0.01}{(0.1)^2}$
$E_{\text{cell}} = 1.56 - 0.02955 \log_{10} \frac{10^{-2}}{10^{-2}}$
$E_{\text{cell}} = 1.56 - 0.02955 \log_{10} (1)$
Since $\log_{10}(1) = 0$, the entire subtracted term becomes zero!
Question: Find the EMF of the following hydrogen gas concentration cell at $298 \text{ K}$:
$Pt | H_2 (P_1 = 2 \text{ atm}) | HCl (0.1 \text{ M}) || HCl (0.1 \text{ M}) | H_2 (P_2 = 0.5 \text{ atm}) | Pt$
Anode (Oxidation): $H_{2(P_1)} \rightarrow 2H^+ + 2e^-$
Cathode (Reduction): $2H^+ + 2e^- \rightarrow H_{2(P_2)}$
Overall: $H_{2(P_1)} \rightarrow H_{2(P_2)}$
Here, $n = 2$.
$E_{\text{cell}} = 0 - \frac{0.0591}{2} \log_{10} \frac{P_2}{P_1}$
$E_{\text{cell}} = -0.02955 \log_{10} \frac{0.5}{2}$
$E_{\text{cell}} = -0.02955 \log_{10} (0.25)$
$\log_{10} (0.25) = \log_{10} (1/4) = -\log_{10}(4) \approx -0.602$
$E_{\text{cell}} = -0.02955 \times (-0.602) \approx +0.0178 \text{ V}$
Question: A standard hydrogen electrode is coupled with another hydrogen electrode immersed in an unknown buffer solution. If the measured EMF of the cell is $0.236 \text{ V}$ at $298 \text{ K}$ and the standard electrode acts as the cathode, calculate the pH of the unknown buffer.
Since SHE is the cathode, the unknown buffer is the anode.
Anode: $H_2 \rightarrow 2{H^+}_{\text{(unknown)}} + 2e^-$
Cathode (SHE): $2{H^+}_{\text{(std)}} + 2e^- \rightarrow H_2$
Overall: $2{H^+}_{\text{(std)}} \rightarrow 2{H^+}_{\text{(unknown)}}$
$n = 2$.
$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log_{10} \frac{[{H^+}_{\text{unknown}}]^2}{[{H^+}_{\text{std}}]^2}$
Since $E^{\circ}_{\text{cell}} = 0$ and $[{H^+}_{\text{std}}] = 1$:
$E_{\text{cell}} = - \frac{0.0591}{2} \times 2 \log_{10} [{H^+}_{\text{unknown}}]$
$E_{\text{cell}} = -0.0591 \log_{10} [{H^+}_{\text{unknown}}]$
Recall that $\text{pH} = -\log_{10}[H^+]$. Therefore:
$E_{\text{cell}} = 0.0591 \times \text{pH}$
$0.236 = 0.0591 \times \text{pH}$
$\text{pH} = \frac{0.236}{0.0591} \approx 3.99$
Question: For the cell $Ag | Ag^+ \text{ (saturated } AgCl \text{ in } 0.1 \text{ M } KCl) || Ag^+ (0.1 \text{ M}) | Ag$, the measured EMF is $0.45 \text{ V}$ at $298 \text{ K}$. Determine the solubility product ($K_{sp}$) of $AgCl$.
Anode: $Ag \rightarrow {Ag^+}_{\text{(anode)}} + e^-$
Cathode: ${Ag^+}_{\text{(cathode)}} + e^- \rightarrow Ag$
Overall: ${Ag^+}_{\text{(cathode)}} \rightarrow {Ag^+}_{\text{(anode)}}$
$n = 1$. $E^{\circ}_{\text{cell}} = 0$.
$E_{\text{cell}} = 0 - \frac{0.0591}{1} \log_{10} \frac{[{Ag^+}_{\text{anode}}]}{[{Ag^+}_{\text{cathode}}]}$
$0.45 = -0.0591 \log_{10} \frac{[{Ag^+}_{\text{anode}}]}{0.1}$
$-7.614 = \log_{10} [{Ag^+}_{\text{anode}}] - \log_{10}(0.1)$
$-7.614 = \log_{10} [{Ag^+}_{\text{anode}}] - (-1)$
$\log_{10} [{Ag^+}_{\text{anode}}] = -8.614$
$[{Ag^+}_{\text{anode}}] = 10^{-8.614} = 2.43 \times 10^{-9} \text{ M}$
In the anode, the solution is $0.1 \text{ M } KCl$, so $[Cl^-] \approx 0.1 \text{ M}$.
$K_{sp} = [Ag^+][Cl^-] = (2.43 \times 10^{-9}) \times (0.1) = 2.43 \times 10^{-10}$
Question: A cell consists of a standard Copper electrode ($Cu^{2+} / Cu$) as the cathode, and a Copper electrode immersed in a solution containing $0.01 \text{ M } Cu^{2+}$ and $1.0 \text{ M } NH_3$ as the anode. If the EMF is $0.38 \text{ V}$, calculate the formation constant ($K_f$) of the complex $[Cu(NH_3)_4]^{2+}$.
This is a concentration cell. $n = 2$. $E^{\circ} = 0$.
$E_{\text{cell}} = - \frac{0.0591}{2} \log_{10} \frac{[{Cu^{2+}}_{\text{free, anode}}]}{[{Cu^{2+}}_{\text{std, cathode}}]}$
$0.38 = -0.02955 \log_{10} \frac{[{Cu^{2+}}_{\text{free, anode}}]}{1}$
$\log_{10} [{Cu^{2+}}_{\text{free, anode}}] = -\frac{0.38}{0.02955} = -12.86$
$[{Cu^{2+}}_{\text{free, anode}}] = 10^{-12.86} = 1.38 \times 10^{-13} \text{ M}$
$Cu^{2+} + 4NH_3 \rightleftharpoons {[Cu(NH_3)_4]^{2+}}$
Initial $Cu^{2+}$ was $0.01 \text{ M}$. Because $K_f$ is usually huge, assume almost all $Cu^{2+}$ converted to the complex. So, $[[Cu(NH_3)_4]^{2+}] \approx 0.01 \text{ M}$.
The concentration of $NH_3$ consumed is $4 \times 0.01 = 0.04 \text{ M}$.
Remaining $[NH_3] = 1.0 - 0.04 = 0.96 \text{ M}$.
$K_f = \frac{[[Cu(NH_3)_4]^{2+}]}{[Cu^{2+}][NH_3]^4}$
$K_f = \frac{0.01}{(1.38 \times 10^{-13})(0.96)^4} = \frac{0.01}{(1.38 \times 10^{-13})(0.849)}$
$K_f = 8.5 \times 10^{10}$
Question: For a reversible cell, $E^{\circ}$ is $1.05 \text{ V}$ at $298 \text{ K}$, and its temperature coefficient $\left( \frac{\partial E}{\partial T} \right)_P$ is $-1.5 \times 10^{-4} \text{ V K}^{-1}$. Calculate the standard enthalpy change ($\Delta H^{\circ}$) for the reaction ($n = 2$).
$E = 1.05 \text{ V}$, $T = 298 \text{ K}$, $n = 2$, $F = 96500 \text{ C mol}^{-1}$.
$\left( \frac{\partial E}{\partial T} \right)_P = -1.5 \times 10^{-4} \text{ V K}^{-1}$.
$\Delta H = -2 \times 96500 \times [1.05 - 298 \times (-1.5 \times 10^{-4})]$
$\Delta H = -193000 \times [1.05 + 0.0447]$
$\Delta H = -193000 \times 1.0947$
$\Delta H = -211277 \text{ J mol}^{-1}$
Question: The EMF of the cell $Pt | H_2 (1 \text{ atm}) | HA (0.1 \text{ M}) || HCl (0.1 \text{ M}) | H_2 (1 \text{ atm}) | Pt$ is $0.118 \text{ V}$ at $298 \text{ K}$. Calculate the degree of dissociation ($\alpha$) of the weak acid $HA$.
$E_{\text{cell}} = -0.0591 \log_{10} \frac{[{H^+}_{\text{anode}}]}{[{H^+}_{\text{cathode}}]}$ (Note: $n=1$ for $H^+ + e^- \rightarrow \frac{1}{2}H_2$, or use $n=2$ with squared terms; result is identical).
$0.118 = -0.0591 \log_{10} \frac{[{H^+}_{\text{anode}}]}{0.1}$
$-2.0 = \log_{10} \frac{[{H^+}_{\text{anode}}]}{0.1}$
$10^{-2} = \frac{[{H^+}_{\text{anode}}]}{0.1}$
$[{H^+}_{\text{anode}}] = 10^{-3} \text{ M}$
For a weak acid, $[H^+] = C \alpha$
$10^{-3} = 0.1 \times \alpha$
$\alpha = 0.01$ (or $1\%$)
Question: The EMF of a cell comprising a Zinc electrode dipped in $0.05 \text{ M } ZnSO_4$ and a Saturated Calomel Electrode (SCE) is $1.08 \text{ V}$ at $298 \text{ K}$. If the reduction potential of the SCE is $+0.244 \text{ V}$, calculate the standard reduction potential of $Zn^{2+}/Zn$.
$E_{\text{cell}} = E_{\text{SCE}} - E_{Zn^{2+}/Zn \text{ (non-standard)}}$
$1.08 = 0.244 - E_{Zn \text{ (non-standard)}}$
$E_{Zn \text{ (non-standard)}} = 0.244 - 1.08 = -0.836 \text{ V}$
$E_{Zn \text{ (non-standard)}} = E^{\circ}_{Zn} - \frac{0.0591}{2} \log_{10} \frac{1}{[Zn^{2+}]}$
$-0.836 = E^{\circ}_{Zn} - 0.02955 \log_{10} \frac{1}{0.05}$
$-0.836 = E^{\circ}_{Zn} - 0.02955 \log_{10} (20)$
$-0.836 = E^{\circ}_{Zn} - 0.02955 \times 1.301$
$-0.836 = E^{\circ}_{Zn} - 0.0384$
$E^{\circ}_{Zn} = -0.836 + 0.0384 = -0.7976 \text{ V}$
Question: For the reaction $Fe^{2+} + Ag^+ \rightleftharpoons Fe^{3+} + Ag$, standard potentials are $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \text{ V}$ and $E^{\circ}_{Ag^+/Ag} = 0.80 \text{ V}$. Calculate the equilibrium constant $K_c$ at $298 \text{ K}$.
Cathode is Silver (higher reduction potential). Anode is Iron.
$E^{\circ}_{\text{cell}} = 0.80 - 0.77 = 0.03 \text{ V}$
Electrons transferred, $n = 1$.
$0.03 = \frac{0.0591}{1} \log_{10} K_c$
$\log_{10} K_c = \frac{0.03}{0.0591} = 0.5076$
$K_c = 10^{0.5076} \approx 3.22$
Question: A Daniel cell ($Zn / Zn^{2+} (1 \text{ M}) || Cu^{2+} (1 \text{ M}) / Cu$) has an initial EMF of $1.10 \text{ V}$. What will be the change in EMF if the anodic half-cell ($Zn^{2+}$) is diluted 100 times?
$E_1 = 1.10 - \frac{0.0591}{2} \log_{10} \frac{1}{1} = 1.10 \text{ V}$
Diluting 100 times means $[Zn^{2+}]_{\text{new}} = 0.01 \text{ M}$.
$E_2 = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.01}{1}$
$E_2 = 1.10 - 0.02955 \times \log_{10}(10^{-2})$
$E_2 = 1.10 - 0.02955 \times (-2) = 1.10 + 0.0591 = 1.1591 \text{ V}$
Change = $E_2 - E_1 = 1.1591 - 1.10 = +0.0591 \text{ V}$.
Question: The standard reduction potential of the quinone-hydroquinone couple ($Q + 2H^+ + 2e^- \rightleftharpoons QH_2$) is $+0.699 \text{ V}$. A platinum electrode is immersed in a solution of pH 5 saturated with quinhydrone. Calculate the reduction potential of this half-cell at 298 K.
$E = E^{\circ} - \frac{0.0591}{2} \log_{10} \frac{[QH_2]}{[Q][H^+]^2}$
Since $[Q] = [QH_2]$, they cancel.
$E = 0.699 - \frac{0.0591}{2} \log_{10} \frac{1}{[H^+]^2}$
$E = 0.699 + \frac{0.0591}{2} \log_{10} [H^+]^2$
$E = 0.699 + 0.0591 \log_{10} [H^+]$
Because $\text{pH} = -\log_{10} [H^+]$:
$E = 0.699 - 0.0591 \times \text{pH}$
$E = 0.699 - 0.0591 \times 5 = 0.699 - 0.2955 = 0.4035 \text{ V}$
Question: A cell $Zn | Zn^{2+} (1 \text{ L}, 0.1 \text{ M}) || Cu^{2+} (1 \text{ L}, 0.1 \text{ M}) | Cu$ operates while delivering a constant current of $9.65 \text{ A}$ for 10 hours. Calculate the EMF of the cell after this time. ($E^{\circ}_{\text{cell}} = 1.10 \text{ V}$).
Charge $Q = I \times t = 9.65 \text{ A} \times (10 \times 3600 \text{ s}) = 347400 \text{ Coulombs}$.
Moles of electrons ($n_e$) = $Q / 96500 = 347400 / 96500 = 3.6 \text{ moles of e}^-$.
Reaction: $Zn \rightarrow Zn^{2+} + 2e^-$. To produce 3.6 moles of $e^-$, 1.8 moles of $Zn^{2+}$ are generated.
New $[Zn^{2+}] = 0.1 \text{ (initial)} + 1.8 = 1.9 \text{ M}$.
Reaction: $Cu^{2+} + 2e^- \rightarrow Cu$. To consume 3.6 moles of $e^-$, 1.8 moles of $Cu^{2+}$ are needed. But wait! The initial amount is only $0.1 \text{ moles}$!
The battery ran out of Copper ions completely well before 10 hours. The $[Cu^{2+}]$ reached zero (or practically zero, chemical equilibrium limit).
Question: Calculate the EMF of $Zn | Zn^{2+} (0.1 \text{ M}) || Cu^{2+} (0.01 \text{ M}) | Cu$ at $350 \text{ K}$. Given $E^{\circ}_{\text{cell}}$ at 298 K is $1.10\text{V}$, and $\Delta S^{\circ} = -20 \text{ J/K}$. Assume $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature.
We know $\Delta S = nF (\frac{\partial E^{\circ}}{\partial T})$.
$-20 = 2 \times 96500 \times \frac{E^{\circ}_{350} - 1.10}{350 - 298}$
$-20 = 193000 \times \frac{E^{\circ}_{350} - 1.10}{52}$
$E^{\circ}_{350} - 1.10 = -0.0054$
$E^{\circ}_{350} = 1.0946 \text{ V}$
Factor = $\frac{2.303 \times 8.314 \times 350}{2 \times 96500} = 0.0347 \text{ V}$
$E = 1.0946 - 0.0347 \log_{10} \frac{0.1}{0.01}$
$E = 1.0946 - 0.0347 \log_{10} (10)$
$E = 1.0946 - 0.0347 = 1.0599 \text{ V}$
Question: Calculate the EMF of the cell: $Zn(Hg) (a_1 = 0.1) | ZnSO_4 (aq) | Zn(Hg) (a_2 = 0.001)$ at 298 K. Here, $a_1$ and $a_2$ represent the activities of Zinc in the mercury amalgam.
Anode: $Zn(Hg)_{a_1} \rightarrow Zn^{2+} + 2e^-$
Cathode: $Zn^{2+} + 2e^- \rightarrow Zn(Hg)_{a_2}$
Overall: $Zn(Hg)_{a_1} \rightarrow Zn(Hg)_{a_2}$
Reaction quotient $Q = \frac{a_2}{a_1}$. $n=2$. $E^{\circ}=0$.
$E = 0 - \frac{0.0591}{2} \log_{10} \frac{0.001}{0.1}$
$E = -0.02955 \log_{10} (0.01)$
$E = -0.02955 \times (-2) = +0.0591 \text{ V}$
Question: Given $E^{\circ}_{Cu^{2+}/Cu^+} = 0.15 \text{ V}$ and $E^{\circ}_{Cu^+/Cu} = 0.50 \text{ V}$. Find the equilibrium constant for the disproportionation reaction: $2Cu^+ \rightleftharpoons Cu + Cu^{2+}$ at 298 K.
Oxidation (Anode): $Cu^+ \rightarrow Cu^{2+} + e^- \quad (E^{\circ}_{\text{ox}} = -0.15 \text{ V})$
Reduction (Cathode): $Cu^+ + e^- \rightarrow Cu \quad (E^{\circ}_{\text{red}} = +0.50 \text{ V})$
Overall: $2Cu^+ \rightarrow Cu + Cu^{2+}$
$E^{\circ}_{\text{cell}} = 0.50 - 0.15 = 0.35 \text{ V}$. $n=1$.
$0.35 = \frac{0.0591}{1} \log_{10} K_c$
$\log_{10} K_c = \frac{0.35}{0.0591} = 5.92$
$K_c = 10^{5.92} \approx 8.3 \times 10^5$
Question: For the cell $Ag | \text{Saturated } Ag_2CrO_4 || Ag^+ (0.1 \text{ M}) | Ag$, $E = 0.165 \text{ V}$. Find the $K_{sp}$ of Silver Chromate ($Ag_2CrO_4$).
$0.165 = -0.0591 \log_{10} \frac{[Ag^+]_{\text{anode}}}{0.1}$
$-2.79 = \log_{10} [Ag^+] - (-1)$
$\log_{10} [Ag^+] = -3.79 \implies [Ag^+] = 1.62 \times 10^{-4} \text{ M}$
Let solubility be $S$. Then $[Ag^+] = 2S$ and $[{CrO_4^{2-}}] = S$.
$2S = 1.62 \times 10^{-4} \implies S = 8.1 \times 10^{-5} \text{ M}$.
$K_{sp} = [Ag^+]^2[{CrO_4^{2-}}] = (2S)^2(S) = 4S^3$
$K_{sp} = 4(8.1 \times 10^{-5})^3 = 2.12 \times 10^{-12}$
Question: Given $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \text{ V}$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \text{ V}$. Calculate $E^{\circ}$ for $Fe^{3+} + 3e^- \rightarrow Fe$.
$\Delta G^{\circ}_1$ for $Fe^{3+} \rightarrow Fe^{2+}$ ($n=1$): $\Delta G^{\circ}_1 = -1 \times F \times 0.77$
$\Delta G^{\circ}_2$ for $Fe^{2+} \rightarrow Fe$ ($n=2$): $\Delta G^{\circ}_2 = -2 \times F \times (-0.44) = +0.88F$
$\Delta G^{\circ}_{\text{total}} = \Delta G^{\circ}_1 + \Delta G^{\circ}_2 = -0.77F + 0.88F = +0.11F$
For the overall reaction ($n=3$):
$\Delta G^{\circ}_{\text{total}} = -3 \times F \times E^{\circ}_{\text{overall}}$
$+0.11F = -3F \times E^{\circ}_{\text{overall}}$
$E^{\circ}_{\text{overall}} = -\frac{0.11}{3} = -0.036 \text{ V}$
Question: The half-reaction is ${MnO_4^-} + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ with $E^{\circ} = 1.51 \text{ V}$. If the pH of the solution is changed from 1 to 3, assuming all other concentrations remain $1 \text{ M}$, by how much does the reduction potential change?
$E = E^{\circ} - \frac{0.0591}{5} \log_{10} \frac{[Mn^{2+}]}{[{MnO_4^-}][H^+]^8}$
Since $[Mn^{2+}] = [{MnO_4^-}] = 1$:
$E = 1.51 - \frac{0.0591}{5} \log_{10} \frac{1}{[H^+]^8}$
$E = 1.51 + \frac{0.0591 \times 8}{5} \log_{10} [H^+]$
$E = 1.51 - 0.09456 \times \text{pH}$
$\Delta E = E_{\text{pH}=3} - E_{\text{pH}=1}$
$\Delta E = (1.51 - 0.09456 \times 3) - (1.51 - 0.09456 \times 1)$
$\Delta E = -0.09456 \times 2 = -0.189 \text{ V}$
Question: Calculate $E_{\text{cell}}$ for $Fe | Fe^{2+} (0.1 \text{ M}) || {MnO_4^-} (0.01 \text{ M}), Mn^{2+} (0.1 \text{ M}), H^+ (0.01 \text{ M}) | Pt$.
Given $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \text{ V}$ and $E^{\circ}_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$.
Oxidation: $(Fe \rightarrow Fe^{2+} + 2e^-) \times 5$
Reduction: $( {MnO_4^-} + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O ) \times 2$
Overall: $5Fe + 2{MnO_4^-} + 16H^+ \rightarrow 5Fe^{2+} + 2Mn^{2+} + 8H_2O$
Total electrons transferred, $n = 10$.
$E^{\circ} = 1.51 - (-0.44) = 1.95 \text{ V}$
$Q = \frac{[Fe^{2+}]^5 [Mn^{2+}]^2}{[{MnO_4^-}]^2 [H^+]^{16}}$
$Q = \frac{(0.1)^5 \times (0.1)^2}{(0.01)^2 \times (0.01)^{16}}$
$Q = \frac{10^{-5} \times 10^{-2}}{10^{-4} \times 10^{-32}} = \frac{10^{-7}}{10^{-36}} = 10^{29}$
$E = 1.95 - \frac{0.0591}{10} \log_{10}(10^{29})$
$E = 1.95 - 0.00591 \times 29 = 1.95 - 0.171 = 1.779 \text{ V}$
Mastering the Fundamentals
These selected problems represent the absolute pinnacle of Class 12 electrochemistry logic. From tracking exact stoichiometric coefficients in the Nernst quotient to navigating the intricate thermodynamics of Latimer diagrams, these methodologies are your key to unlocking the highest percentiles in JEE Advanced.
As you practice further, always write out the balanced half-reactions first. Over $70\%$ of calculation errors in Nernst equation problems stem from incorrect '$n$' values or missed exponent powers in the Reaction Quotient ($Q$). For more foundational theory, always refer back to our core resources on Derivations and Fundamentals.
