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{cu}^{2+} + 2\text{e}^-
ightarrow \text{cu} ) | +0.34 |
| ( \text{cr}^{3+} + 3\text{e}^-
ightarrow \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 species with the more positive reduction potential will be reduced (cathode), and the other will be oxidized (anode). Since \( \text{Cu}^{2 + }/ \text{Cu} \) has a reduction potential of \( + 0.34\space V \), and the cell is spontaneous (\( E_{\text{cell}}=1.07\space V>0 \)), \( \text{Cu}^{2+} \) will be reduced (cathode) and \( \text{Cr} \) will be oxidized (so the \( \text{Cr}^{3+}/\text{Cr} \) half - reaction will be reversed, meaning we use the reduction potential of \( \text{Cr}^{3+}/\text{Cr} \) as the anode's reduction potential in the formula).

The formula for \( E_{\text{cell}} \) is \( E_{\text{cell}}=E_{\text{red (cathode)}}-E_{\text{red (anode)}} \). Here, \( E_{\text{red (cathode)}} = E_{\text{red (Cu)}}= + 0.34\space V \), \( E_{\text{cell}} = 1.07\space V \), and \( E_{\text{red (anode)}}=E_{\text{red (Cr)}} \) (the reduction potential of chromium we need to find).

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

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

Substitute the known values: \( E_{\text{red (Cr)}}=0.34 - 1.07 \)

Step3: Calculate the value

\( 0.34-1.07=- 0.73 \)

Answer:

\(-0.73\)