Question

5. In aqueous solution Cuions are more stable than Cu ions because 2 Cu" (aq) + Cu (aq) + Cu(s) (a) Give an oxidation hall-cell reaction and a reduction half-cell reaction that can be combined to form this overall reaction. (0.5 marks) (b) Use Eq. 18 with your answer to (a) to calculate the theoretical K for the reaction. (1 mark) 6. Refer to Fig. 1. The salt bridge is to prevent wholesale mixing of the two solutions. Assume that the bridge is removed and that wholesale mixing of the two solutions occurred as shown here: ZnSO. (IMM CuSO. (1M) (a) What value should the voltmeter show? (0.5 mark) (b) There is a reaction. Give an equation for it and specify where it occurs. (1 mark)

Answer 1

5.

a)

Oxidation reaction ( anode)

Cu⁺ (aq)   $\to$ Cu²⁺ (aq) ; E^o(Cu²⁺/Cu⁺) = 0.15 V

Reduction reaction (cathode)

Cu⁺ (aq)  $\to$ Cu (s) ; E^o(Cu⁺/Cu) = 0.52 V

E^o_cell = E^o_cathode - E^o_anode = 0.52 - 0.15 = 0.37 V.

b)

Number of electrons transferred (n) = 1.

E^o_cell = log K.

0.0592

Or, 0.37 = 0.0592 × log K.

Or, logK = (0.37/0.0592)

Or, log K = 6.25

Or, K = 10^6.25 = 1.78×10⁶ .

6.

Concentration of both the solution is 1.0 M.

Then, E_cell = E^o_cell .

E^o_cell of given daniell cell = 1.1 V

Now, if salt bridge is removed there will be a charge imbalance hence voltage shown in voltmeter will be zero.

b.

Anode Reaction

Zn - 2e $\to$ Zn²⁺ (aq)

Cathode reaction.

Cu²⁺ (aq) + 2e $\to$ Cu

Zn + CuSO₄ $\to$    ZnSO₄ + Cu .

But in absence of salt bridge the reaction will stop because electrical neutrality is absent.