Electrochemical cells (1) - Standard electrode potentials.
(Use Section 1 of the IB Chemistry data booklet for any relevant equations and Section 24 to obtain the
relevant E⦵ values where they are not given).
1. Consider a cell made up from the following two half-cells:
Ni2+(aq) + 2e– ⇌ Ni(s) E⦵= – 0.26 V
Ag+(aq) + e– ⇌ Ag(s) E⦵ = + 0.80 V
(a) Deduce the spontaneous reaction that will occur when the cell is operating.
(b) Identify the positive and negative electrodes of the cell.
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(c) Identify the species that is being oxidised and the species that is being reduced.
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(d) Deduce the overall cell potential under standard conditions.
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(e) Calculate the Gibbs free energy change for this reaction at standard conditions.
2. Draw a labelled diagram of a cell made up from a Mg(s)/Mg2+(aq) half-cell connected to a
Mn(s)/Mn2+(aq) half-cell operating under standard conditions.
Show the direction of electron flow in the external circuit and deduce the potential of this cell in volts.
3. Deduce the equation for the spontaneous reaction that occurs when a dichromate half-cell, Cr2O72(aq),H+(aq)/Cr3+(aq),H2O(l), comes into contact with an iodine half-cell, ½I2(l)/I–(aq).
4. Use standard redox potentials to show whether tin(II) ions can reduce iron(III) ions.
(Given: Sn4+(aq)/Sn2+(aq) E⦵ = + 0.15 V)
5. Show that when copper(I) salts come into contact with water they undergo disproportionation. That
is, they are spontaneously oxidized and reduced at the same time.
Answers
1. (a) Ni(s) + 2Ag+(aq) →Ni2+(aq) + 2Ag(s)
(b) Positive electrode: silver; Negative electrode: nickel (as the nickel half-cell has a more
negative E⦵ value than the silver half-cell)
(c) Nickel metal is oxidized and silver ions are reduced.
(d) 1.06 V (being the difference between – 0.26 V and + 0.80 V)
(e) ΔG = − 2 (mol) x 96500 (C mol-1) x 1.06 (V )= − 204580 J = − 205 kJ
2.
Cell potential = 1.18 V (the difference between -2.37 V and -1.19 V)
3. Cr2O72-(aq) + 6I–(aq) + 14H+(aq) → 2Cr3+(aq) + 3I2(aq) + 7H2O(l)
(Acidified dichromate ions are a stronger oxidizing agent than iodine so oxidize iodide ions to iodine)
4. Sn4+(aq)/Sn2+(aq) ││ Fe3+(aq)/Fe2+(aq)
E⦵ = + 0.15 V E = + 0.77 V
Electrons flow from the Sn4+(aq)/Sn2+(aq) half-cell to the Fe3+(aq)/Fe2+(aq) half-cell so Sn2+(aq) can reduce
Fe3+(aq) and the spontaneous reaction is :
Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq) E
cell=
0.62 V
5. Cu(s)/Cu+(aq) ││Cu+(aq) / Cu2+(aq)
E⦵ = + 0.52 V E⦵ = + 0.15 V
Electrons flow from Cu+(aq) / Cu2+(aq) to Cu(s)/Cu+(aq)
Half-equations: Cu+(aq) → Cu2+(aq) + e– and Cu+(aq) + e– → Cu(s)
Overall redox equation: 2Cu+(aq) → Cu2+(aq) + Cu(s) E⦵cell= 0.37 V
so the oxidation number of copper changes from +1 to 0 and +2, i.e. Cu+(aq) disproportionates.