A voltaic cell is constructed with two Zn^2+ -Zn electrodes. The two cell compon

Question

A voltaic cell is constructed with two Zn^2+ -Zn electrodes. The two cell components have [Zn^2+] = 1.7 M and [Zn^2+] = 1.00 Times 10^-2 M, respectively. What is the standard emf of the cell? What is the cell emf of the concentration given? Express your answer using two significant figures. For the anode component predict whether |Zn^2+| will increase, decrease, or stay the same as the cell operates. For the cathode component predict whether |Zn^2+| will increase, decrease, or stay the same as the cell operates.

Explanation / Answer

1)

we know that

for a cell

Eo = Eo cathode - Eo anode

in this case

both the electrodes are zinc

so

Eo cathode = Eo anode

Eo cell = Eo cathode - Eo anode= 0

so

the value of Eo cell is 0

2)

now

according to nernst equation

E = Eo - (0.05916/n) log Q

now

anode reaction :

Zn (s) ---> Zn+2 (anode) + 2e-

cathode reaction :

Zn+2 (cathode) + 2e- ---> Zn (s)

the overall reaction is

Zn (s) + Zn+2 (cathode) ---> Zn+2 (anode) + Zn (s)

the reaction quotient is given by

Q = [Zn+2 (anode) ] /[Zn+2 (cathode)]

using the given values

Q = [1 x 10-2 ] / [1.7]

Q = 5.88235 x 10-3

now

E = Eo - (0.05916/n) log Q

here

n =2 as two electrons are transferred

so

E = 0 - (0.05916/2) log 5.88235 x 10-3

E = 0.066 V

so

emf of the cell is 0.066 V

3)

oxidation takes place at anode

so

the anode reaction is

Zn (s) ---> Zn+2 + 2e-

so

as the cell operates

[Zn+2] will increase

4)

reduction takes place at cathode

so

the reaction is

Zn+2 + 2e- --> Zn

so

as the cell operates

[Zn+2] will decrease

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