Some reactions of chromium ions are shown below.

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 E^o 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 E^o value is more negative than of Cr⁺³ to Cr⁺²]

ii)

M1: value of E_cell correctly calculated (with correct sign) for metal named in (i)

M2: E^o _cell is positive and so reaction is feasible

[E_cell is calculated by subtracting lower value (oxidation) from a greater value(reduction)]

(b) M1: (Cr₂O₇ ^2– + 14H⁺ + 6e^– ⇌ 2Cr³⁺ + 7H₂O) E^o = +1.33V

 (H₂O₂ + 2H⁺ + 2e^– ⇌ 2H₂O) E^o = +1.77V E^o _cell = 0.44 V

M2: E^o cell (0.44V) is positive (so the reaction is feasible)/E^o (Cr₂O₇ ^2– /Cr³⁺) is less positive than E^o (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