Correct answer Carries: 4.
Wrong Answer Carries: -1.
A cell has \( E^\circ_{cell} = 0.52 \, V \) and an equilibrium constant of \( 10^{17} \) at 298 K. How many electrons are transferred?
\( E^\circ_{cell} = \frac{0.059}{n} \log K_c \), \( 0.52 = \frac{0.059}{n} \times 17 \), \( n = \frac{1.003}{0.52} \approx 2 \).
A conductivity cell with a 0.015 M KNO\(_3\) solution has a resistance of 300 Ω and a cell constant of 0.75 cm\(^{-1}\). What is the molar conductivity?
\( \kappa = \frac{\text{cell constant}}{R} = \frac{0.75}{300} = 0.0025 \, S \, cm^{-1} \).
\( \Lambda_m = \frac{\kappa \times 1000}{c} = \frac{0.0025 \times 1000}{0.015} = 166.67 \, S \, cm^2 \, mol^{-1} \).
A cell \( Cd(s) | Cd^{2+}(0.005 \, M) || Br_2(l) | Br^-(0.01 \, M) | Pt(s) \) operates at 298 K. What is the cell potential? (Given: \( E^\circ_{Cd^{2+}/Cd} = -0.40 \, V \), \( E^\circ_{Br_2/Br^-} = 1.07 \, V \))
\( E^\circ_{cell} = 1.07 - (-0.40) = 1.47 \, V \).
\( E_{cell} = 1.47 - \frac{0.059}{2} \log \frac{[Cd^{2+}][Br^-]^2}{1} = 1.47 - 0.0295 \log (0.005 \times 0.0001) = 1.47 + 0.103 = 1.573 \, V \).
What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr\(_2\)(SO\(_4\))\(_3\) solution using a current of 0.5 A? (Atomic mass of Cr = 52 g/mol, F = 96500 C/mol)
\( Cr^{3+} + 3e^- \rightarrow Cr \). 1 mol Cr (52 g) requires 3F.
Moles = \( \frac{0.585}{52} = 0.01125 \, mol \), Charge = \( 0.01125 \times 3 \times 96500 = 3256.875 \, C \).
\( t = \frac{Q}{I} = \frac{3256.875}{0.5} = 6513.75 \, s \).
How many electrons are involved in the reduction of 1 mole of \( SO_4^{2-} \) to \( S \) in acidic medium?
\( SO_4^{2-} + 8H^+ + 6e^- \rightarrow S + 4H_2O \). 6 electrons per mole.
In a fuel cell, what is the electrolyte commonly used?
Fuel cells often use KOH as the electrolyte in alkaline conditions.
What is the emf of the cell \( Co(s) | Co^{2+}(0.02 \, M) || H^+(0.01 \, M) | H_2(g)(1 \, bar) | Pt(s) \) at 298 K? (Given: \( E^\circ_{Co^{2+}/Co} = -0.28 \, V \), \( E^\circ_{H^+/H_2} = 0.00 \, V \))
\( E^\circ_{cell} = 0.00 - (-0.28) = 0.28 \, V \).
\( E_{cell} = 0.28 - \frac{0.059}{2} \log \frac{0.02}{0.01^2} = 0.28 - 0.07375 = 0.20625 \, V \).
What is the emf of the cell \( Sn(s) | Sn^{2+}(0.05 \, M) || H^+(0.1 \, M) | H_2(g)(1 \, bar) | Pt(s) \) at 298 K? (Given: \( E^\circ_{Sn^{2+}/Sn} = -0.14 \, V \), \( E^\circ_{H^+/H_2} = 0.00 \, V \))
\( E^\circ_{cell} = 0.00 - (-0.14) = 0.14 \, V \).
\( E_{cell} = 0.14 - \frac{0.059}{2} \log \frac{0.05}{0.1^2} = 0.14 + 0.039 = 0.179 \, V \).
During electrolysis of aqueous K\(_2\)SO\(_4\) with inert electrodes, 0.336 L of gas (STP) is collected at both electrodes. How many coulombs were passed? (F = 96500 C/mol)
Cathode: \( 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \), Anode: \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \).
Total moles = \( \frac{0.336}{22.4} = 0.015 \, mol \) (H\(_2\):O\(_2\) = 2:1), H\(_2\) = 0.01 mol, O\(_2\) = 0.005 mol.
Charge = \( (0.01 \times 2 + 0.005 \times 4) \times 96500 = 0.04 \times 96500 = 3860 \, C \).
How many Faradays are required to deposit 1.62 g of silver from AgNO\(_3\) solution? (Atomic mass of Ag = 108 g/mol)
\( Ag^+ + e^- \rightarrow Ag \). 1 mol Ag (108 g) requires 1F.
Moles = \( \frac{1.62}{108} = 0.015 \, mol \), Charge = \( 0.015 \times 1 = 0.015 \, F \).
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