Problem: An electrochemical cell (battery) consists of a Cu plate immersed in 100 mL of 0.055 M Cu2+ solution and a Zn plate immersed in 100 mL of 0.550 M Zn2+ solution. The two compartments are connected by a salt bridge and the cell is maintained at 25 °C. The cell is discharged by passing a 10.00 mA current for 105 seconds.The standard reduction potentials are given below:Cu2+ (aq) + 2 e- → Cu (s)              ε°red = + 0.34 VZn2+ (aq) + 2 e- → Zn (s)               ε°red = - 0.76 VWhat is the change of the potential (voltage) of the cell caused by the discharge? Please circle your answer.

FREE Expert Solution

We are asked to calculate the change of the potential (voltage) of the cell caused by the discharge.


We are given the following values based on the previous items: 

[Zn2+]final = 0.602 M

[Cu2+]final = 0.003 M

Ecell initial = 1.07 V

cell = 1.10 V

n = 2 e-


We will use the Nernst Equation to calculate the cell potential. The Nernst Equation relates the concentrations of compounds and cell potential.


Ecell =E°cell - (0.0592 Vn)logQ

Ecell = cell potential under non-standard conditions
cell = standard cell potential
n = number of e- transferred
Q= reaction quotient = [products]/[reactants] 



We go through the following steps to solve the problem: 

Step 1. Calculate the final cell potential (Ecell)
Step 2. Calculate the change in voltage

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Problem Details

An electrochemical cell (battery) consists of a Cu plate immersed in 100 mL of 0.055 M Cu2+ solution and a Zn plate immersed in 100 mL of 0.550 M Zn2+ solution. The two compartments are connected by a salt bridge and the cell is maintained at 25 °C. The cell is discharged by passing a 10.00 mA current for 105 seconds.

The standard reduction potentials are given below:

Cu2+ (aq) + 2 e- → Cu (s)              ε°red = + 0.34 V

Zn2+ (aq) + 2 e- → Zn (s)               ε°red = - 0.76 V

What is the change of the potential (voltage) of the cell caused by the discharge? Please circle your answer.