Question

The electrochemical cell reaction of the Daniel cell is 

Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu_(s) 

What is the change in the cell voltage on increasing the ion concentration in the anode compartment by a factor 10?

Answer 1

Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu_(s) 

ln the case E°_cell = 1.1V

Reaction quotient Q for the above reaction is,Q =

If suppose concentration of Cu²⁺ is 1 .OM then the concentration of Zn²⁺ is 10M (why because, ion concentration in the anode compartment increased by a 10 factor)

= 1.1 – 0.02955 ……….(1) 

= 1.070 V (cell voltage decreased) 

Thus, the initial voltage is greater than E° because Q < 1. 

As the reaction proceeds, [Zn²⁺] in the anode compartment increases as the zinc electrode dissolves, while [Cu²⁺] in the cathode compartment decreases as metallic copper is deposited on the electrode.

During this process, the Q = [Zn²⁺] [Cu²⁺] steadily increases and the cell voltage therefore steadily decreases. [Zn²⁺] will continue to increase in the anode compartment and [Cu²⁺] will continue to decrease in the cathode compartment. Thus the value of Q will increase further leading to a further decrease in value.