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:

[Zn^{2+}]_{final} = 0.602 M

[Cu^{2+}]_{final} = 0.003 M

E_{cell initial} = 1.07 V

E°_{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.

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}\mathbf{}\mathbf{V}}{\mathbf{n}}\mathbf{\right)}{\mathbf{logQ}}}$

E_{cell} = cell potential under non-standard conditions

E°_{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 (E _{cell})*

An electrochemical cell (battery) consists of a Cu plate immersed in 100 mL of 0.055 M Cu^{2+} solution and a Zn plate immersed in 100 mL of 0.550 M Zn^{2+} 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 10^{5} seconds.

The standard reduction potentials are given below:

Cu^{2+} (aq) + 2 e^{-} → Cu (s) ε°_{red} = + 0.34 V

Zn^{2+} (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.

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