Question

a spontaneous cell made from copper and chromium has an ( e_{\text{cell}} ) value of 1.07 v.

half - reaction	( e_{\text{red}} ) (v)
( \text{cr}^{3+}+3\text{e}^-\to\text{cr} )	?

calculate the value of the reduction potential for the chromium.
( e_{\text{red}} ) of ( \text{cr} = ? ) v
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for cell potential

For a spontaneous cell, \( E_{\text{cell}} = E_{\text{cathode (reduction)}} - E_{\text{anode (reduction)}} \). A spontaneous cell has a positive \( E_{\text{cell}} \). We need to determine which half - reaction is reduction (cathode) and which is oxidation (anode). The half - reaction with the more positive reduction potential will be the reduction (cathode) reaction. The reduction potential of \( \ce{Cu^{2+}/Cu} \) is \( + 0.34\space V \), and we need to find the reduction potential of \( \ce{Cr^{3+}/Cr} \). Since the cell is spontaneous, the copper half - reaction (with higher reduction potential) will be the reduction (cathode) and the chromium half - reaction will be the oxidation (anode, so we will use the reverse of its reduction half - reaction for oxidation). So the formula becomes \( E_{\text{cell}}=E_{\text{red (cathode)}}-E_{\text{red (anode)}} \), where cathode is \( \ce{Cu^{2+}/Cu} \) and anode is \( \ce{Cr^{3+}/Cr} \) (but we use its reduction potential in the formula).

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

We know \( E_{\text{cell}} = 1.07\space V \) and \( E_{\text{red (cathode)}}= + 0.34\space V \). From \( E_{\text{cell}}=E_{\text{red (cathode)}}-E_{\text{red (anode)}} \), we can rearrange to \( E_{\text{red (anode)}}=E_{\text{red (cathode)}} - E_{\text{cell}} \)

Step3: Substitute the values into the formula

Substitute \( E_{\text{cell}} = 1.07\space V \) and \( E_{\text{red (cathode)}} = 0.34\space V \) into the formula:
\( E_{\text{red (Cr^{3+}/Cr)}}=0.34 - 1.07=- 0.73\space V \)

Answer:

\(-0.73\)