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CHEM//37

Electron Transfer Exam Question 2020/21

Notes in purple are additional notes on typical responses / errors

o- Lead,acid batteries are a common form of rechargeable battery- The electrode

reactions for the cell are shown in Equations (o) and (2)- The electrolyte is concentrated sulfuric acid either in liquid or gel form-

Pbo2(s)＋H2so4(aq) ＋＋2H＋(aq)＋2e, A Pbso4(s) ＋2H2o Eo = o-698 V (2)

(a) Determine the overall cell reaction+ standard cell potential and Gibbs energy per

mol of Pb for this battery- [3]

Pbo2 ＋Pb＋2H2so4 A 2Pbso4＋2H2o

Ecell = Efinal · Einitial = o-698 · (,/-332) = 2-/3 V

2 e transferred per mol Pb: /G = ,nF/E = ,2 x 96485 x 2-/3 = ,39o-7 kJ mol,o Pb

(o mark) (o mark)

(o mark)

Common error: Using n = o instead of n = 2 i-e- not calculating for o mol Pb- some students also missed negative sign-

Use the Nernst equation for Equation (2) to explain why the voltage of the battery declines during use- Assuming a constant current is drawn4 from the battery

describe the time dependence of this voltage decline- [6]

E = Eo ＋ (RT.nF) ln ([o].[R])

= o-698 ＋(RT.2F) ln ([Pbo2][H2so4][H＋]2.[Pbso4] [H2o]2) (o mark)

Can approach explanation for voltage decline in a couple of ways:

Assuming [Pbo2] = [Pbso4] = [H2o] = o as solid or in excess

E = o-698＋ (RT.2F) ln ([H2so4][H＋]2)

so as sulphuric acid concentration decreases (is used up) the value of E will decrease and hence cell potential declines-

or can use full eq: E = o-698＋ (RT.2F) ln ([Pbo2][H2so4][H＋]2.[Pbso4] [H2o]2) [Pbo2] and [H2so4] will go down with time and [Pbso4] will go up · again leads to

decrease in E with time and hence decrease in Ecell- (either explanation = 2 marks)

Constant current means that the concentrations of the reactants will decrease linearly with time- The ln dependence of E on conc means the time dependence will be of a logarithmic nature- so very slow decline initially followed by a much quicker decline as

conc needs to drop many orders of magnitude before effect is seen for E- (3 marks)

Most students addressed the first 3 marks well (using a range of valid approaches)- Fewer students were awarded the next 3 marks · most could say that current decreased over time+ but few made the link between constant current and constant decrease in concentration and then the log nature of the relationship with E-

(c) A cyclic voltammogram (CV) was recorded for the reaction in Eq (o) as shown below- The CV starts at ,/-2 V and sweeps first towards more negative potentials-

(i) Assign the reactions responsible for peaks A and B and propose the reaction that results in increased current C- [3]

A = reduction = Pbso4(s) ＋2H＋(aq) ＋2e, A Pb(s) ＋H2so4(aq)

B = oxidation = Pb(s) ＋H2so4(aq) A Pbso4(s) ＋2H＋(aq) ＋2e,

C = proton reduction = 2H＋＋2e, A H2 (also accept water reduction)

(o mark per assignment)

Most students could assign A and B- Reaction C was less well addressed- I also gave /-5 mark if C assigned to ‘reduction of background electrolyte二 but with no species named-

(ii) sketch the shape of the CV expected for the reversible reduction of a

dissolved redox species under diffusion,control and explain why the CV has this form- Hence explain why peaks A and B look different from

those obtained for diffusing species- [7]

sketch: expecting ox and red peaks of approx- equal magnitude separated by

approx- 59.n mV+

(o mark)

Explanation: o) Initial increase in current due to overcoming activation barrier to ET as predicted by Butler,Volmer (ET kinetics controlled current)- 2) Eventually ET is so fast that diffusion of fresh reactant cannot keep up with rate of ET- 3) Current therefore reaches max (peak) and then drops as controlled by rate of mass

transport (diffusion)- (3 marks · o per sentence)

students can of course use different phrasing but I was look for the three concepts: o- ET controlled region+ 2- diffusion controlled region and 3- why there is a peak-

A and B reactions are not purely diffusion,controlled as limiting reactants are solid electrodes- Initial increase in current due to kinetics as described in point o) above but maximum and current decrease caused by electrode surface having fully reacted rather than slow supply of fresh reactant by diffusion- Leads to sharper peaks and more rapid fall in current than predicted from diffusion control-

(3 marks · o per sentence)

This is meant to be more challenging as is unseen · but better students answered this successfully- Many students talked (incorrectly) instead about reaction being quasi or irreversible-

(d) Explain the following observations:

(i)over,charging this battery can lead to water loss and hydrogen and oxygen gas production- [3]

Charging of battery is opposite of cell reaction: 2Pbso4＋2H2o A Pbo2 ＋Pb＋2H2so4

(omark)

side reaction from over,charging (once Pbso4 all converted) is water splitting- Leading to o2 and H2 products:

At positive electrode: H2o A - o2 ＋2e, ＋2H＋

At negative electrode: 2H2o＋2e, A H2 ＋2oH,

(o mark)

(o marks)

The reactions were in many cases correctly identified (or partly identified) but many

answers did not address what the charging process was and why

overcharging would lead to these reactions-