3. (10 marks) The electrochemical cell Zn(s) | Zn2+ (? M) 11 Ni2+ (1.00 M) | Ni(

Question

3. (10 marks) The electrochemical cell Zn(s) | Zn2+ (? M) 11 Ni2+ (1.00 M) | Ni(s) is constructed using a completely immersed zinc electrode that weighs 32.68 g and a nickel electrode immersed in 575 mL of 1.00 M Ni2+ (aq) solution. A steady current of 0.0715 A is drawn from the cell as the electrons move from the zine electrode to the nickel electrode. Ni2+ (aq) + 2e-_? Ni(s) E" =-0.23 V a. Which reactant is the limiting reactant in this cell? b. How long, in hours, does it take for the cell to be completely discharged? c. What mass has the nickel electrode gained when the cell is completely discharged? Il is completely dischar

Explanation / Answer

a)

Overall reaction is

Zn(s) + Ni2+(aq) <-----> Ni(s) + Zn2+(aq)

stoichiometrically, 1mole of Zn(s) reacting with 1mole of Ni2+

No of moles of Zn = 32.68g/65.38g/mol = 0.49985

No of moles of Ni2+ = (1.00mol/1000ml)×575ml = 0.575mol

Therefore,

Limiting reagent is Zn

b)

1coulomb in 1second = 1A

0.0715 coulomb in 1second = 0.0715A

1mole of Zn give 2moles of electron

0.49985moles of Zn give 0.9997moles of electrons

No of coulomb for 1mole of electron = 96485coulomb

So,

No of coulomb for 0.9997moles of electrons = (96485coulomb/1mol)×0.9997mol = 96456coulomb

time required to completely discharged = 96456/0.0715 =1349034seconds/3600 = 374.7hours

c)

No of moles of Zn dissolved = 0.49985

No of moles of Ni priduced= 0.49985

Molar mass of Ni = 58.69g/mol

Mass of Ni gained = 58.69g/mol ×0.49985mol = 29.34g

d) Initial moles of Ni2+ = 0.575

moles of Ni2+ consumed = 0.49985

remaining moles of Ni2+ = 0.575 - 0.49985 =0.07515

Concentration of Ni2+ after complete discharge  = (0.07515mol/575ml)×1000ml =0.1307M

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