Q24. Given the standard reduction potentials, E°Ag+/Ag = 0.80 V and E°Mg2+/Mg = -2.37 V
E° for Ag+/Ag(0.1 M) + Mg(s) → Ag(s) + Mg2+(aq, 0.2 M)
at 25°C (K° = ?) (round off to two decimal places).
(Given: Faraday constant = 96500 C mol-1, Gas constant R = 8.314 J K-1 mol-1)
The cell potential for the given electrochemical cell is calculated using standard reduction potentials and the Nernst equation at 25°C.
Problem Breakdown
The half-reactions are:
Cathode (reduction): \ce{MnO4- (aq) + 8H+ (aq) + 5e- -> Mn^{2+} (aq) + 4H2O (l)}, E0 = -2.37 V
Anode (oxidation): \ce{Fe2+ (aq) -> Fe3+(aq) + e-}, E0ox = -0.80 V (reverse of given E0red = +0.80 V for \ceFe3+/Fe2+)
Given conditions: [MnO4-] = 1 M, [Mn2+] = 0.1 M, Faraday constant F = 96500 C/mol, R = 8.314 J K-1 mol-1.
Cell Reaction and E°cell
Balanced cell reaction (multiply anode by 5):
\ce{MnO4-+ 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O}
E0\ce{cell} = E0\ce{cathode} - E0\ce{anode} = -2.37 - (0.80) = -3.17 V
Nernst Equation Calculation
At 298 K (25°C): E\ce{cell} = E0\ce{cell} - RT/nF.ln Q, where n = 5.
2.303 RT/F = 0.0591 V (at 25°C).
Q = [\ceMn2+] [\ceFe3+]5/[\ceMnO4-] [\ceFe2+]5 [\ce{H+}]8}
Assuming standard [H+] = 1 M, [Fe3+] = 1 M, [Fe2+] = 1 M (as unspecified): Q = 0.1.
ln(0.1) = -2.303, so -0.0591/5 × (-2.303) = +0.0272 V.
E\ce{cell} = -3.17 + 0.0272 = -3.143 V ≈ -3.14 V (2 decimal places).
cell potential MnO4- Fe2+ electrochemical cell
The cell potential MnO4- Fe2+ electrochemical cell is a key topic in electrochemistry for CSIR NET Life Sciences aspirants. This guide explains the calculation using given standard reduction potentials: E0\ceMnO4-/Mn2+= -2.37 V and E0\ceFe3+/Fe2+= 0.80 V at 25°C, with concentrations [MnO4-] = 1 M, [Mn2+] = 0.1 M.
Step-by-Step Solution
- Identify cathode (more negative E°): MnO4- reduction despite negative value (cell spontaneous if E_cell > 0, but here negative indicates non-spontaneous under standard conditions).
- Balance electrons: 5 Fe2+ oxidized per MnO4-.
- Compute
E0\cecell = -2.37 - 0.80 = -3.17 V. - Apply Nernst:
E = -3.17 -0.0591/5.log(0.1) = -3.17 + 0.027 = -3.14 V.
Common Exam Options Explained
- -3.17 V: Ignores Nernst correction for [Mn2+] = 0.1 M.
- -2.37 V: Only cathode potential.
- -3.14 V: Correct, includes Q = 0.1 effect (positive shift).
- 0 V: Assumes equilibrium (incorrect).
This cell potential MnO4- Fe2+ electrochemical cell problem tests Nernst application in acidic medium, vital for CSIR NET. Practice with Faraday constant (96500 C/mol) and R (8.314 J/mol K) for Gibbs energy links.
