An electrochemical cell is constructed with a zinc metal anode in contact with a

Question

An electrochemical cell is constructed with a zinc metal anode in contact with a 0.020 M solution of zinc(II) nitrate and a silver cathode in contact with a 0.0010 M solution of silvrr(I) nitrate. What is the emfof this cell at 15degreeC? (F = 96500 C/mol electrons, R = 8.3145 J/(K-mol) and T in K) E_celldegree = E_reddegree (cathode) - E_reddegree (anode) 2. Copper metal is purified by electrolysis. How much copper metal could be produced from copper(II) oxide by applying a current of 8.0 amps at the appropriate negative potential for 56.0 hours? (Amp = C/t) 3. Iodine-131 has a half-life of 8.1 days and is used as a tracer for the thyroid gland. If a patient drinks a sodium iodide (NaI) solution containing iodine-131 on a Tuesday, how many days will it take for the concentration of iodine-131 to drop to 10.0% of its initial concentration? E_cell = E_cell - RT/nF InQ In N_t/N_0 = -0.693 t/t_1/2

Explanation / Answer

We have standard reduction potential as,

Zn2+ (aq.) + 2e- ------------> Zn (s)         E0(Zn2+/Zn) = -0.76 V

Ag+ (aq.) + 1e- -------------> Ag (s)         E0(Ag+/Ag) = + 0.80 V

Hence oxidation will occur at Zn (i.e. an anode) and reduction will occur at Ag electrode (i.e. Cathode)

E0cell = E0 (cathode) – E0 (anode)

E0cell = E0(Ag+/Ag) - E0(Zn2+/Zn)

E0cell = (+0.80) – (-0.76)

E0cell = 0.80 + 0.76

E0cell = + 1.56 V

Hence balanced redox reaction is,

Zn (s) + 2Ag+ (aq.) ----------> Zn2+ (aq.) + 2Ag

This is a 2e- transfer reaction and hence for this reaction,

n = 2

Nernst equation is given as,

Ecell = E0cell – (RT/nF) ln([Products]/[reactants])

For above cell,

Ecell = E0cell – (RT/nF) ln([Zn2+]/[Ag+]2) ………. (1)

Solids have activity = 1 and hence concentrations of solid is 1

For given cell, Ecell = ? at,

T = 15 0C = 15 + 273.15 = 288.15 K

n = 2

F = 96500 C/mol electrons

R = 8.3145 J/(K.mol)

[Zn2+] = [Zn(NO3)2] = 0.020 M

[Ag+] = [AgNO3] = 0.0010 M

Let us put all these values in above Nernst equation (1)

Ecell = 1.56 – (8.314 x 288.15/2 x 96500) ln{(0.02)/(0.0010)2}

Ecell = 1.56 – (0.0124) (9.903)

Ecell = 1.56 – 0.12

Ecell = 1.44 V

EMF of the cell is +1.44 V.

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