Question

Question 2. [10 Marks] Given the following half-cell reduction potentials: Ni2+ (aq) + 2e-F Ni(s), Ered = -0.23 V Pt2+ (aq) + 2e-E Pt(s), Ered = +1.2 V Pd2+(aq) + 2e- EPd(s), Ered = 0.99 V a) Sketch the cell for which the overall cell potential is the greatest. [3 Marks) b) Identify the cathode and anode, and then show the direction of electron flow for the cell in the circuit (a). [3 Marks] c) Will Pt(s) reduce Pd2+(aq)? Explain your answer. [2 marks] d) What will happen if a strip of Ni(s) is placed into a solution containing Pt2+(aq)? Explain your answer. [2 marks]

Answer 1

This question requires high analysis using only one basic point, so let us answer it in step by step manner:

Basic point needed here:

1) Greater the reduction potential more easily the substance is reduced and hence acts as strong oxidising agent.

Smaller the reduction potential more easily the substance is oxidized and hence acts as strong reducing agent

2) At anode always oxidation reaction takes place and at cathode always reduction takes place.

In a simple manner :

If you arrange the given data, as numbers in a number line form, using Electrode potential value of cell; the top one(having more negative value) will be the strong oxidising agent, acts as the anode and the bottom one(having more positive value) will be strong reducing agent,acts as an cathode.

Another point is that the top one displace the down one, because the top one is strong oxidizing agent, that means it oxidizes (loss of electrons) the down one and itself undergoes reduction(that elctrons will be used by this)

example: Zn + CuSO₄ -----> Cu + ZnSO₄ (Zn is top of Cu, in electrochemical series)

Zn + Increasing strength of oxidising agent Standard reduction potentials at 298 K (Electrochemical Series) Reduction half Standard reduction reaction potential E° (in volts) Litté L - 3.05 K+té -2.93 Ba2+ + 2e Ba -2.90 Ca2+ + 2e Са -2.87 Na+ + é Na -2.71 Mg2+ + 2e Mg -2.37 A13+ + 3e ΑΙ – 1.66 1 Mn2+ + 2e Mn - 1.18 2H200 + 2e H2(g) + 20H (aq) -0.83 Zn2+ + 2e -0.76 Cr3+ + 3e ar -0.74 Fe2+ + 2e Fe -0.44 Cd2+ + 2e Cd -0.40 o Co2+ + 2e Co -0.28 Ni2+ + 2e Ni - 0.25 Sn2+ + 2e Sn -0.14 Pb2+ + 2e Pb -0.13 2H+ + 2e H -0.00 Sn4++ 2e +0.15 Cu2+ + e Cu +0.15 Cut + e Cu +0.52 12 +2e 21- +0.53 Fe3+ + Fe2+ +0.77 Ag+ té Ag +0.80 Br2+2e 2Br +1.08 Cl2+2e 2C1 +1.36 O2(g) + 4H* (aq) +4€ 2H2O +1.23 Au3+ + 3e Au +1.50 Co3+ té Co2+ +1.82 F₂ +2e 2F +2.87 ↑ ↑ ↑ ↑ Î Î Î Î Î Î Î Î Î ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ Increasing strength of reducing agent → Sn2+

Arranging the given question in order of electrochemical series, we get;

Ni⁺²_(aq) + 2 e^- <==> Ni_(s) E⁰_red = -0.23V

Pd⁺²_(aq) + 2 e^- <==> Pd_(s)    E⁰_red = +0.99V

Pt⁺²_(aq) + 2 e^- <==> Pt_(s) E⁰_red = +1.2V

Using the above points you can say that only three cells can be formed

1) Ni-Pd cell

2)Ni-Pt cell

3)Pd-Pt cell (All are in anode-cathode format)

for Ni-Pd cell E⁰_red = E⁰_cathode - E⁰_anode

= 0.99 -(-0.23)

= 1.22 V

for  Ni-Pt cell   E⁰_red = E⁰_cathode - E⁰_anode

   = 1.20 - (-0.23)

= 1.43 V

for  Pd-Pt cell   E⁰_red = E⁰_cathode - E⁰_anode

= 1.20 - (0.99)

= 0.41 V

In this all, the greatest  E⁰_red is having for  Ni-Pt cell.

(a)

>e G Conventional Current(I) direction Ni rod. Salt bridge →pt rod 는 Ni +2 Anode Pt+3 Cathode (reduction half-cell) (oxidation half-cell) It has two half-cells Ni - +2 Ni + Që Eo = -0.23 V (At anode) Eo=+1.2v Pt+2+2 é → pt (At Cathode)

b) As it is clearly mentioned in the question,

cathode is Pt(Platinum)    

anode is Ni(Nickel)

and conventional current direction is from cathode ( Platinum) to anode (Nickel)

(c) No, as Pt⁺² has more oxidizing character than Pd⁺² (from first figure we can say that),as Pt⁺² has more oxidizing character, it always wanted to transform into Pt than Pd⁺².

(d)It is like

Zn + CuSO₄ -----> Cu + ZnSO₄  

hence , the top element displaces the bottom element from its solution

Therefore, Ni displaces Pt from the solution.