Problem: Given the voltaic cell below, find the cell potential.Zn(s), [Zn2+]=4.50M] || Cu(s), [Cu2+]=0.0120MCu2+(aq) + 2e– ⇌ Cu(s)       E° = 0.34 VZn2+(aq) + 2e– ⇌ Zn(s)       E° = -0.76 VA) 1.23 VB) 1.10 VC) 1.18 VD) 1.02 VE) 1.08 V

FREE Expert Solution

We are asked to calculate for the cell potential (Ecell) for the voltaic cell below

Zn(s), [Zn2+] = 4.50M] || Cu(s), [Cu2+] = 0.0120M

We can rewrite the cell notation as:

Zn(s) | Zn2+, (4.50 M) || Cu2+, (0.0120M) | Cu(s)

Recall that the Nernst Equation relates the concentrations of compounds and cell potential.

Ecell=E°cell-(0.05916 Vn) log Q

Ecell = cell potential under non-standard conditions
cell = standard cell potential
n = mole e- transferred
Q = reaction quotient = products/reactants

We're going to calculate for the Ecell using the following steps:

Step 1: Identify the anode and the cathode in the reaction and write the overall reaction
Step 2: 
Calculate the cell potential of the reaction.
Step 3: 
Calculate Ecell using the Nernst Equation.

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Problem Details

Given the voltaic cell below, find the cell potential.

Zn(s), [Zn2+]=4.50M] || Cu(s), [Cu2+]=0.0120M

Cu2+(aq) + 2e ⇌ Cu(s)       E° = 0.34 V

Zn2+(aq) + 2e ⇌ Zn(s)       E° = -0.76 V

A) 1.23 V

B) 1.10 V

C) 1.18 V

D) 1.02 V

E) 1.08 V

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