Question

A voltaic cell is constructed that uses the following reaction and operates at 298 K. Zn(s) + Ni2+(aq) Zn2+(aq) + Ni(s) (a)

What is the emf of this cell under standard conditions?______ V

(b) What is the emf of this cell when [Ni2+] = 2.24 M and [Zn2+] = 0.112 M?_____ V

(c) What is the emf of the cell when [Ni2+] = 0.298 M and [Zn2+] = 0.925 M?____ V

Answer 1

Ni2+(aq) + 2 e --> Ni(s) here Eo = –0.257
Zn(s) --> Zn2+(aq) + 2 e here Eo = +0.763

so
Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)
has Eo = 0.506 volts & n = 2 moles of electrons

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nernst equation is
E = Eo - (0.0592 /n) (log Q)
E = Eo - (0.0592 /n) (log Q)
E = 0.506 - (0.0592 / 2) (log [products] / reactants]
E = 0.506 - (0.0296) (log [Zn+2] / [Ni+2]

a) What is the emf of this cell when [Ni2+ ] = 2.24 M and [Zn2+ ] = 0.112 M?
E = 0.506 - (0.0296) (log [Zn+2] / [Ni+2]
E = 0.506 - (0.0296) (log [0.112]/ [2.24]
E = 0.506 - (0.0296) (log 0.047)
E = 0.506 - (0.0296) (-1.32)
E = 0.506+ 0.039
your first answer is
E = 0.545 volts
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(b) What is the emf of the cell when [Ni2+ ] = 0.298 M and [Zn2+ ] = 0.925M?
E = 0.506 - (0.0296) (log [Zn+2] / [Ni+2]
E = 0.506 - (0.0296) (log [0.925] / [0.298]
E = 0.506 - (0.0296) (log 3.10)
E = 0.506 - (0.0296) (0.491)
E = 0.506 - 0.014
your secoond answer is
E = 0.42 volts