Question

half-reaction | e° (v)
e²⁺ + 2e⁻→ e | + 0.76
g⁺ + e⁻ → g | + 0.10
j²⁺ + 2e⁻→ j | 0.00
l⁺ + e⁻ → l | - 0.82
what is the e°_cell value for the non - spontaneous reaction between g and l?
+0.92 v
+0.72 v
-0.92 v
-0.72 v

Explanation:

Step1: Identify Oxidation and Reduction

For a non - spontaneous reaction, the cell potential \(E^{\circ}_{cell}\) is negative. First, we need to determine which species is oxidized and which is reduced. The half - reactions are:

• Reduction: \(G^{+}+e^{-}\to G\), \(E^{\circ}_{red}= + 0.10\ V\)
• Oxidation: \(L\to L^{+}+e^{-}\) (reverse of \(L^{+}+e^{-}\to L\)), so \(E^{\circ}_{ox}=-E^{\circ}_{red}\) of \(L^{+}+e^{-}\to L\). The \(E^{\circ}_{red}\) for \(L^{+}+e^{-}\to L\) is \(- 0.82\ V\), so \(E^{\circ}_{ox}=+ 0.82\ V\)

Step2: Calculate \(E^{\circ}_{cell}\)

The formula for \(E^{\circ}_{cell}\) is \(E^{\circ}_{cell}=E^{\circ}_{red}+E^{\circ}_{ox}\) (for the overall reaction). But wait, actually, for the overall reaction, if we have reduction of \(G^{+}\) and oxidation of \(L\), the overall reaction is \(G^{+}+L\to G + L^{+}\).
The correct formula is \(E^{\circ}_{cell}=E^{\circ}_{red}(cathode)-E^{\circ}_{red}(anode)\). In a non - spontaneous reaction, the cathode (reduction) and anode (oxidation) are reversed from the spontaneous case.
Wait, let's do it properly. The standard cell potential \(E^{\circ}_{cell}=E^{\circ}_{cathode}(reduction)-E^{\circ}_{anode}(reduction)\). For a non - spontaneous reaction, \(E^{\circ}_{cell}<0\).
If we consider the reaction between \(G\) and \(L\), the possible reactions:
Case 1: If \(G\) is reduced and \(L\) is oxidized:
Reaction: \(G^{+}+L\to G + L^{+}\)
\(E^{\circ}_{cell}=E^{\circ}_{red}(G^{+}\to G)-E^{\circ}_{red}(L^{+}\to L)\)
\(E^{\circ}_{red}(G^{+}\to G)= + 0.10\ V\), \(E^{\circ}_{red}(L^{+}\to L)=- 0.82\ V\)
\(E^{\circ}_{cell}=0.10-(- 0.82)=0.92\ V\) (spontaneous, since \(E^{\circ}_{cell}>0\))
Case 2: If \(G\) is oxidized and \(L\) is reduced:
Reaction: \(G + L^{+}\to G^{+}+L\)
\(E^{\circ}_{cell}=E^{\circ}_{red}(L^{+}\to L)-E^{\circ}_{red}(G^{+}\to G)\)
\(E^{\circ}_{red}(L^{+}\to L)=- 0.82\ V\), \(E^{\circ}_{red}(G^{+}\to G)= + 0.10\ V\)
\(E^{\circ}_{cell}=-0.82 - 0.10=-0.92\ V\) (non - spontaneous, since \(E^{\circ}_{cell}<0\))

Answer:

-0.92 V