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

For a spontaneous electrochemical cell, \( E_{\text{cell}} = E_{\text{red (cathode)}} - E_{\text{red (anode)}} \). Since Al is oxidized, it is the anode (oxidation occurs at anode), and Sn is reduced (reduction occurs at cathode). So \( E_{\text{red (cathode)}} \) is the reduction potential of Sn, and \( E_{\text{red (anode)}} \) is the reduction potential of Al (but since it's oxidized, we use its reduction potential in the formula as the anode's reduction potential).
Given \( E_{\text{cell}} = 1.52 \, \text{V} \), \( E_{\text{red (Sn)}} = -0.14 \, \text{V} \), and we need to find \( E_{\text{red (Al)}} \).

Step2: Rearrange the formula to solve for \( E_{\text{red (Al)}} \)

From \( E_{\text{cell}} = E_{\text{red (cathode)}} - E_{\text{red (anode)}} \), we can rearrange it as \( E_{\text{red (anode)}} = E_{\text{red (cathode)}} - E_{\text{cell}} \).
Substituting the known values: \( E_{\text{red (Al)}} = E_{\text{red (Sn)}} - E_{\text{cell}} \)
Plugging in the numbers: \( E_{\text{red (Al)}} = -0.14 \, \text{V} - 1.52 \, \text{V} \)

Step3: Calculate the result

\( -0.14 - 1.52 = -1.66 \, \text{V} \)

Answer:

\( -1.66 \)