Problem: What is the Ecell for Mg(s) I Mg2+(aq)(0.15 M) II Au+(aq)(0.3 M) I Au(s) ? Assume that the number of electrons transferred in the reaction is n = 2.Mg2+ + 2e- → Mg                 E°red = -2.36 VAu+ + e- → Au                      E°red = +1.69 V1. 4.04 V2. 4.08 V3. 3.2 x 10-2 V4. 4.03 V5. 4.01 V

FREE Expert Solution

We are asked to calculate for the Ecell for the reaction

Mg(s) I Mg2+(aq)(0.15 M) II Au+(aq)(0.3 M) I Au(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

What is the Ecell for Mg(s) I Mg2+(aq)(0.15 M) II Au+(aq)(0.3 M) I Au(s) ? Assume that the number of electrons transferred in the reaction is n = 2.
Mg2+ + 2e- → Mg                 E°red = -2.36 V
Au+ + e- → Au                      E°red = +1.69 V

1. 4.04 V

2. 4.08 V

3. 3.2 x 10-2 V

4. 4.03 V

5. 4.01 V

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