The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn 2 + ( aq , ? M ) ∣ ∣ Zn ( s ) is 25.0 mV at 25 °C . Calculate the concentration of the Zn 2 + ( aq ) ion at the cathode.

The concentration of ion at cathode is 0.704 M

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  

Reduction half reaction (cathode):  

In this case, the cathode and anode both are same. So, will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

where,

n = number of electrons in oxidation-reduction reaction = 2

= 25.0 mV = 0.025 V   (Conversion factor: 1 V = 1000 mV)

= ? M

= 0.100 M

Putting values in above equation, we get:

Hence, the concentration of ion at cathode is 0.704 M

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"