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   $E = 0.06 V$

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

     $E = E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}$

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

So

      $E = 0 - \frac{0.0591}{2} * log[\frac{ 2.00*10^{-2}}{1.6} ]$

=>      $E = 0.06 V$