Arrange the following cells in order of their decreasing EMF values
Given: \( E^{o}_{\mathrm{H^+|H_2}}=0,\ E^{o}_{\mathrm{Cu^{2+}|Cu}}=0.34\ \mathrm{V},\ E^{o}_{\mathrm{Zn^{2+}|Zn}}=-0.76\ \mathrm{V} \) and \( E^{o}_{\mathrm{Fe^{2+}|Fe}}=-0.44\ \mathrm{V} \)
A. Pt(s)|H2 (g, 1 bar)|H+ (aq, 1 M)||Cu2+ (aq, 1 M)|Cu(s)
B. Pt(s)|H2 (g, 1 bar)|H+ (aq, 1 M)||H+ (aq, 0.1 M)|H2 (g, 1 bar) |Pt
C. Zn(s)|Zn2+ (aq, 0.1 M)||Cu2+ (aq, 1 M)|Cu(s)
D. Fe(s)|Fe2+(0.001 M)||H+ (aq, 1 M)|H2 (g, 1 bar)| Pt(s)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C, D, A, B
The EMF of a cell at 298 K is given by the Nernst equation:
\( \displaystyle E_{cell}=E^{\circ}_{cell}-\frac{0.059}{n}\log Q,\quad E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode} \)
Q has the product ion concentrations on top and the reactant ion concentrations below. The left electrode of each cell is the anode.
A. All species are standard, so Ecell = E°cell:
\( \displaystyle E_A=0.34-0=0.34\ \mathrm{V} \)
B. Both electrodes are hydrogen electrodes, so E°cell = 0 (a concentration cell). \( \ce{H^{+}} \) is formed at 1 M (anode) and used up at 0.1 M (cathode):
\( \displaystyle \ce{2H^+}(0.1\ \mathrm{M}) \rightarrow \ce{2H^+}(1\ \mathrm{M}) \)
\( \displaystyle E_B=0-\frac{0.059}{2}\log\frac{(1)^2}{(0.1)^2}=-\frac{0.059}{2}\times2=-0.059\ \mathrm{V} \)
C. \( \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \), n = 2:
\( \displaystyle E^{\circ}_{cell}=0.34-(-0.76)=1.10\ \mathrm{V} \)
\( \displaystyle E_C=1.10-\frac{0.059}{2}\log\frac{0.1}{1}=1.10+0.0295\approx1.13\ \mathrm{V} \)
D. \( \ce{Fe(s) + 2H^+(aq) -> Fe^2+(aq) + H2(g)} \), n = 2:
\( \displaystyle E^{\circ}_{cell}=0-(-0.44)=0.44\ \mathrm{V} \)
\( \displaystyle E_D=0.44-\frac{0.059}{2}\log\frac{0.001}{(1)^2}=0.44+0.0885\approx0.53\ \mathrm{V} \)
Shortcut: C has by far the largest E°cell and B, a concentration cell, is near zero. Options (3) and (4) both start with C and end with B; D (E° = 0.44 V, raised further by the dilute \( \ce{Fe^{2+}} \)) is above A (0.34 V), which leaves Option (3).