For a given galvanic cell, the standard cell potential can be calculated by subtracting the standard half-cell potential of the reaction that occurs at the cathode from the standard half-cell potential of the reaction that occurs at the anode: e∘cell=e∘red(reduction process)−e∘red(oxidation process)

for a given galvanic cell, the standard cell potential can be calculated by subtracting the standard half-cell potential of the reaction that occurs at the cathode from the standard half-cell potential of the reaction that occurs at the anode:

e∘cell=e∘red(reduction process)−e∘red(oxidation process)

part a

calculate the standard cell potential given the following standard reduction potentials:
al3++3e−→al; e∘=−1.66 v
cu2++2e−→cu; e∘=0.340 v
express your answer to two decimal places and include the appropriate units.

E°cell= 2.00 V

Explanation:

Let's consider the following reductions with their respective standard reduction potentials.

Al³⁺ + 3 e⁻ → Al; E°red= −1.66 V

Cu²⁺ + 2 e⁻ → Cu; E°red= 0.340 V

The one with the higher standard reduction potential will occur as a reduction (cathode) while the other will occur as an oxidation (anode). The corresponding half-reactions are:

Anode (oxidation): Al → Al³⁺ + 3 e⁻; E°red= −1.66 V

Cathode (reduction): Cu²⁺ + 2 e⁻ → Cu; E°red= 0.340 V

The standard cell potential (E°cell) can be calculated by subtracting the standard half-cell potential of the reaction that occurs at the cathode from the standard half-cell potential of the reaction that occurs at the anode:

E°cell=E°red(reduction process)−E°red(oxidation process) = 0.340V - (-1.66V) = 2.00V

h let me think

away from the curb

explanation:

the wheels would then hit the curb if it started to rolling, decreasing the chance of an accident.