electrochemical exam questions bank
WITH ANSWERS
1. A disproportionation response happens when a species M+ spontaneously undergoes
simultaneous oxidation and discount.
Beneath consists of E facts for copper and mercury species.
Cu2+(aq) + e− → Cu+(aq) + zero.15
Cu+(aq) + e− → Cu(s) + 0.Fifty two
Hg2+(aq) + e− → Hg+(aq) + zero.91
Hg+(aq) + e− → Hg(l) + 0.Eighty
Using these facts, which one of the following may be predicted?
A Both Cu(I) and Hg(I) undergo disproportionation. B Only Cu(I) undergoes
disproportionation.
C Only Hg(I) undergoes disproportionation.
D Neither Cu(I) nor Hg(I) undergoes
( - ANSWER-b
A lead-acid cellular may be recharged.
2. Write an equation for the overall reaction that happens when the mobile is being
recharged?
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- → PbSO4(s) + 2H2O
(I) +1.Sixty nine
PbSO4(s) + H+(aq) + 2e- → Pb(s) + HSO4-(aq)
to be calculated - ANSWER-Opposite for reduction
(PbS04) --> Same side = add 2 together
2PbS04 ==> for HS04-
2PbSO4 + 2H2O → Pb +PbO2 + 2HSO4- + 2H+
3. A rechargeable nickel-cadmium mobile is an opportunity
, Cd(OH)2(s) + 2e- ----> Cd(s) + 2OH-(aq) -zero.88
NiO(OH)(s) + H2O(I) + e- ---> Ni(OH)2(s) + OH- zero.52
Deduce the oxidation country of the nickel on this cell after recharging is whole. Write an
equation for the overall response that takes place while the cell is recharged. -
ANSWER-commonly nickel will be reduced
in recharging = opposite response = so oxidation will take place
consequently O.S = +3 in NiO(OH)
eq=
contrary
(+ve) = oxidised
acidification,
25.Zero cm3 of a solution of hydrogen peroxide
reacted precisely with 16.2 cm3 of a 0.0200 mol dm-3 answer of potassium
manganate(VII).
4. Basic equation for the reaction is given under.
2Mn04- + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
(i) Use eq for reaction to decide
conc, in g dm-three, of hydrogen peroxide answer? - ANSWER-moles of h202 =
reacting ratio
awareness = moles / volume * a thousand
(moldm-3)
gdm-three = mol dm-three × Mr
Ag+(aq) + e− --> Ag(s) (+0.Eighty)
Fe3+(aq) + e− --> Fe2+(aq) (+0.Seventy seven)
Fe2+(aq) + 2e− ---> Fe(s) (-0.Forty four)
5. Use statistics from table to predict +
provide an explanation for redox reactions
arise when iron powder (s) added to extra of aqueous silver nitrate - ANSWER-Eϴ Ag+
> Eϴ Fe2+
Eϴ Ag+ > Eϴ Fe3+ --> (careful)
WITH ANSWERS
1. A disproportionation response happens when a species M+ spontaneously undergoes
simultaneous oxidation and discount.
Beneath consists of E facts for copper and mercury species.
Cu2+(aq) + e− → Cu+(aq) + zero.15
Cu+(aq) + e− → Cu(s) + 0.Fifty two
Hg2+(aq) + e− → Hg+(aq) + zero.91
Hg+(aq) + e− → Hg(l) + 0.Eighty
Using these facts, which one of the following may be predicted?
A Both Cu(I) and Hg(I) undergo disproportionation. B Only Cu(I) undergoes
disproportionation.
C Only Hg(I) undergoes disproportionation.
D Neither Cu(I) nor Hg(I) undergoes
( - ANSWER-b
A lead-acid cellular may be recharged.
2. Write an equation for the overall reaction that happens when the mobile is being
recharged?
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- → PbSO4(s) + 2H2O
(I) +1.Sixty nine
PbSO4(s) + H+(aq) + 2e- → Pb(s) + HSO4-(aq)
to be calculated - ANSWER-Opposite for reduction
(PbS04) --> Same side = add 2 together
2PbS04 ==> for HS04-
2PbSO4 + 2H2O → Pb +PbO2 + 2HSO4- + 2H+
3. A rechargeable nickel-cadmium mobile is an opportunity
, Cd(OH)2(s) + 2e- ----> Cd(s) + 2OH-(aq) -zero.88
NiO(OH)(s) + H2O(I) + e- ---> Ni(OH)2(s) + OH- zero.52
Deduce the oxidation country of the nickel on this cell after recharging is whole. Write an
equation for the overall response that takes place while the cell is recharged. -
ANSWER-commonly nickel will be reduced
in recharging = opposite response = so oxidation will take place
consequently O.S = +3 in NiO(OH)
eq=
contrary
(+ve) = oxidised
acidification,
25.Zero cm3 of a solution of hydrogen peroxide
reacted precisely with 16.2 cm3 of a 0.0200 mol dm-3 answer of potassium
manganate(VII).
4. Basic equation for the reaction is given under.
2Mn04- + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
(i) Use eq for reaction to decide
conc, in g dm-three, of hydrogen peroxide answer? - ANSWER-moles of h202 =
reacting ratio
awareness = moles / volume * a thousand
(moldm-3)
gdm-three = mol dm-three × Mr
Ag+(aq) + e− --> Ag(s) (+0.Eighty)
Fe3+(aq) + e− --> Fe2+(aq) (+0.Seventy seven)
Fe2+(aq) + 2e− ---> Fe(s) (-0.Forty four)
5. Use statistics from table to predict +
provide an explanation for redox reactions
arise when iron powder (s) added to extra of aqueous silver nitrate - ANSWER-Eϴ Ag+
> Eϴ Fe2+
Eϴ Ag+ > Eϴ Fe3+ --> (careful)
