Question 5
(a) Some reactions of chromium ions are shown below.
Reference: Past Exam Paper – 9701 March 2016 Paper 42 Q5
Solution:
(a)
i) any metal with an Eo value more negative than
–0.41V, e.g. Fe, Mn, Zn, Mg, Cr, Al
[Observing the net charge on the complex ion, we can calculate the charge
on the Cr ion in both complexes. We see that charge reduces from +3 to +2 and
hence we need a metal that will itself go through oxidation and act as a
reducing agent. Therefore we need a metal whose Eo value is more
negative than of Cr+3 to Cr+2]
ii)
M1: value of Ecell correctly calculated (with
correct sign) for metal named in (i)
M2: Eo cell is positive and so
reaction is feasible
[Ecell is calculated by subtracting lower value (oxidation)
from a greater value(reduction)]
(b) M1: (Cr2O7 2–
+ 14H+ + 6e– ⇌ 2Cr3+ + 7H2O) Eo
= +1.33V
(H2O2
+ 2H+ + 2e– ⇌
2H2O) Eo = +1.77V Eo cell
= 0.44 V
M2: Eo cell (0.44V) is positive (so the reaction
is feasible)/Eo (Cr2O7 2– /Cr3+)
is less positive than Eo (H2O2 /H2O)
(c) M1: Cr2O7 2–:
ox.no Cr = +6 because –2 = 2 × ox.no(Cr) + (7 × –2) CrO4 2–:
ox.no Cr = +6 because –2 = ox.no(Cr) + (4 × –2) M2: no change in oxidation
number, so reaction is not redox
[calculate oxidation numbers of Cr in both ionic states and calculate
change in oxidation number(final – initial);
positive change=> oxidation
negative change => reduction
no change=> not a redox reaction]
(d) M1: no. moles Cr deposited = 0.0312/ 52 =
6.0 × 10–4 moles
M2: deduction that 6 moles of e– needed per mole
of Cr/ reaction is Cr2O7 2- + 14H+
+ 12e– → 2Cr + 7H2O
[since there is no electrode reaction given for Cr2O7
2- to Cr(metal), we need to combine two separate reactions to
achieve: Cr2O7 2- + 14H+ + 12e–
→ 2Cr + 7H2O.
2 moles of chromium metal requires 12 moles of electrons as seen from above
equation, therefore one mole of metal requires 6 moles of electrons. ]
M3: no. moles of e– = 6 × 6.0 × 10–4
= (0.125 × t)/ 96 500 so t = (6 × 6.0 × 10–4 × 96 500)/(0.125 × 60)
= 46.3min/ 0.772 h/ 2780s
Solutions provided by Kashish Varshney, India
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