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 cell potential formula

For a spontaneous electrochemical cell, \( E_{\text{cell}} = E_{\text{cathode (reduction)}} - E_{\text{anode (reduction)}} \). Since Al is oxidized, it is the anode (oxidation occurs at anode), and Sn is reduced (cathode, as reduction occurs at cathode). So \( E_{\text{cell}} = E_{\text{red (Sn)}} - E_{\text{red (Al)}} \).

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

We have \( E_{\text{red (Al)}} = E_{\text{red (Sn)}} - E_{\text{cell}} \).

Step3: Substitute the given values

Given \( E_{\text{cell}} = 1.52 \, \text{V} \) and \( E_{\text{red (Sn)}} = -0.14 \, \text{V} \). Substituting these values: \( E_{\text{red (Al)}} = -0.14 - 1.52 \).

Step4: Calculate the result

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

Answer:

-1.66