Question

What is the cell emf for the concentrations given? Express your answer using two significant figures.

Answer 1

Complete Question

A voltaic cell is constructed with two
`Zn^(2+)`
`- Zn` electrodes. The two cell compartment have
`[Zn^(2+)] =  1.6 \ M` and
`[Zn^(2+)] =  2.00*10^(-2) \  M` respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

The value is
`E =  0.06 V`

Step-by-step explanation:

Generally from the question we are told that

The concentration of
`[Zn^(2+)]` at the cathode is
`[Zn^(2+)]_a =  1.6 \ M`

The concentration of
`[Zn^(2+)]` at the anode is
`[Zn^(2+)]_c =  2.00*10^(-2) \  M`

Generally the the cell emf for the concentration is mathematically represented as

`E =  E^o - (0.0591)/(2) log([Zn^(2+)]a)/( [Zn^(2+)]c)`

Generally the
`E^o`is the standard emf of a cell, the value is 0 V

So

`E =  0  -  (0.0591)/(2)  * log[( 2.00*10^(-2))/(1.6) ]`

=>
`E =  0.06 V`