A cell is constructed using Cu^ 2+ Cu and Al^ 3+ Al electrodes. 1. Identify cathode and anode. 2. Write the reactions at anode, cathode and net cell reaction. 3. Calculate E^0_cell E^0_ Al^ 3+ Ag =-1.66V E^0_ Cu^ 2+ Cu =+0.34V

It is measured by potentiometer which does not have internal resistance. 1. 'Al' will act as anode, 'Cu' will act as cathode because reduction potential of Cu^ 2+ Cu is higher. 2. 3. E^0_cell=E^0_ cu^ 2+ Cu -E^0_ Al^ 3+ Al =+0.34V-(-1.66V) =2.00V

A cell is constructed using Cu^ 2+ Cu and Al^ 3+ Al electrodes. 1…

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Question

A cell is constructed using $\frac{\text{Cu}^{2+}}{\text{Cu}}$ and $\frac{\text{Al}^{3+}}{\text{Al}}$ electrodes.

Identify cathode and anode.
Write the reactions at anode, cathode and net cell reaction.
Calculate $\text{E}^0_\text{cell}$

$\text{E}^0_\frac{\text{Al}^{3+}}{\text{Ag}}=-1.66\text{V}$

$\text{E}^0_\frac{\text{Cu}^{2+}}{\text{Cu}}=+0.34\text{V}$

✓

Answer

It is measured by potentiometer which does not have internal resistance.

'Al' will act as anode, 'Cu' will act as cathode because reduction potential of $\frac{\text{Cu}^{2+}}{\text{Cu}}$ is higher.

$\text{E}^0_\text{cell}=\text{E}^0_\frac{\text{cu}^{2+}}{\text{Cu}}-\text{E}^0_\frac{\text{Al}^{3+}}{\text{Al}}$

$=+0.34\text{V}-(-1.66\text{V})$

$=2.00\text{V}$

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