A voltaic cell is constructed using solutions of NaHSO4, H2SO3, and MnSO4 with suitable electrodes. — Electrochemistry Chemistry Question
Problem Context
A voltaic cell is constructed using solutions of NaHSO4, H2SO3, and MnSO4 with suitable electrodes. The relevant half reactions are:
HSO4 – (aq) + 3H+ (aq) + 2e– H2SO3 (aq) + H2O E˚ = 0.17 V
Mn 2+(aq) + 2e – Mn(s) E˚ = –1.18 V
Sketch a working voltaic cell on which you
i. identify the contents of each half-cell,
ii. suggest a substance to be used as an electrode in the HSO4 – (aq)/H2SO3(aq) half-cell.
iii. label the anode and cathode,
iv. show the direction of electron movement in the external circuit,
v. indicate the direction of movement of cations in the salt bridge.
Model Answer
Mn cations e– Pt/C anode cathode
Mn2+ HSO4¯
H2SO3
Write the equation for the overall cell reaction and calculate the cell potential at standard conditions.
Model Answer
Mn + HSO4 – + 3 H+ Mn2+ + H2SO3 + H2O
Σ ocell = Σ oox + Σ ored Σ ocell = 1.18 + 0.17 = 1.35 V
Calculate the cell potential if the [H + ] is changed to pH = 1.
Model Answer
Σ = 1.35 – 0.089 = 1.26 V
Predict the qualitative effect on the cell potential if a solution of Ba(NO3)2 is added to both cell compartments. Justify your prediction.
[BaSO3 Ksp = 8.3 10 –7 ; BaSO4 Ksp = 1.2 10 –10]
Model Answer
Ba(NO3)2 will have no effect on the anode components. Ba(NO3)2 will react with both H2SO3 and HSO4 – in cathode compartment but BaSO4 is less soluble so reaction will shift to the left, decreasing the potential.
Predict the qualitative effect on the cell potential if the electrode that serves as the anode is doubled in size. Justify your prediction.
Model Answer
There is no effect on the cell potential by changing the size of the anode. Only the solution component concentrations and gas pressures affect E.