# 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 initial potential (voltage) of the electrochemical cell? Please circle your answer.

###### FREE Expert Solution

We’re being asked to calculate the initial potential (voltage) of the electrochemical cell.

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

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. Write the two half-cell reactions
Step 2. Identify the oxidation half-reaction (anode) and the reduction half-reaction (cathode)
Step 3Determine the half-cell potentials (refer to the Standard Reduction Potential Table)
Step 4. Calculate E°cell (standard).
Step 5. Calculate Ecell. ###### 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