Complete Question
A voltaic cell is constructed with two [tex]Zn^{2+}[/tex][tex]- Zn[/tex] electrodes. The two cell compartment have [tex][Zn^{2+}] = 1.6 \ M[/tex] and [tex][Zn^{2+}] = 2.00*10^{-2} \ M[/tex] respectively.
What is the cell emf for the concentrations given? Express your answer using two significant figures
Answer:
The value is [tex]E = 0.06 V[/tex]
Explanation:
Generally from the question we are told that
The concentration of [tex][Zn^{2+}][/tex] at the cathode is [tex][Zn^{2+}]_a = 1.6 \ M[/tex]
The concentration of [tex][Zn^{2+}][/tex] at the anode is [tex][Zn^{2+}]_c = 2.00*10^{-2} \ M[/tex]
Generally the the cell emf for the concentration is mathematically represented as
[tex]E = E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}[/tex]
Generally the [tex]E^o[/tex]is the standard emf of a cell, the value is 0 V
So
[tex]E = 0 - \frac{0.0591}{2} * log[\frac{ 2.00*10^{-2}}{1.6} ][/tex]
=> [tex]E = 0.06 V[/tex]
