Question

al is oxidized in a spontaneous cell made from tin and aluminum. the overall cell has an ( e_{\text{cell}} ) value of 1.52 v. the reduction potential of tin is -0.14 v. calculate the value of the reduction potential for the aluminum. ( e_{\text{red}} ) of al = ? v enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for cell potential

The formula for the cell potential \( E_{\text{cell}} \) is \( E_{\text{cell}} = E_{\text{cathode (reduction)}} - E_{\text{anode (reduction)}} \). Since Al is oxidized, it is the anode (oxidation occurs at the anode), and Sn is reduced (reduction occurs at the cathode). So, \( E_{\text{cathode (Sn)}} = - 0.14 \, \text{V} \), \( E_{\text{cell}} = 1.52 \, \text{V} \), and we need to find \( E_{\text{anode (Al)}} \) (which is the reduction potential of Al, \( E_{\text{red (Al)}} \)).

Step2: Rearrange the formula to solve for \( E_{\text{anode (reduction)}} \)

From \( E_{\text{cell}} = E_{\text{cathode (reduction)}} - E_{\text{anode (reduction)}} \), we can rearrange it to \( E_{\text{anode (reduction)}} = E_{\text{cathode (reduction)}} - E_{\text{cell}} \).

Step3: Substitute the known values

Substitute \( E_{\text{cathode (reduction)}} = - 0.14 \, \text{V} \) and \( E_{\text{cell}} = 1.52 \, \text{V} \) into the formula:
\( E_{\text{red (Al)}} = - 0.14 - 1.52 \)
\( E_{\text{red (Al)}} = - 1.66 \, \text{V} \)

Answer:

-1.66