Question

question 13 of 30
the half - reaction at the anode in an electrochemical cell is given below. what other half - reaction would most likely occur at the cathode to produce a nonspontaneous reaction?
ag(s) → ag⁺(aq)+e⁻
click for a reduction potential chart
a. ni²⁺(aq)+2e⁻ → ni(s)
b. zn(s) → zn²⁺(aq)+2e⁻
c. ag⁺(aq)+e⁻ → ag(s)
d. hg²⁺(aq)+2e⁻ → hg(l)

Explanation:

Step1: Recall the condition for non - spontaneous reaction

For a non - spontaneous reaction in an electrochemical cell, $E_{cell}<0$. The anode reaction is $Ag(s)\to Ag^{+}(aq)+e^{-}$, with $E^{\circ}_{Ag^{+}/Ag} = 0.80\ V$. The cathode reaction is a reduction. The cell potential is given by $E_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$.

Step2: Analyze each option

• Option A: For $Ni^{2 + }(aq)+2e^{-}\to Ni(s)$, $E^{\circ}_{Ni^{2+}/Ni}=- 0.25\ V$. Then $E_{cell}=E^{\circ}_{Ni^{2+}/Ni}-E^{\circ}_{Ag^{+}/Ag}=-0.25 - 0.80=-1.05\ V<0$.
• Option B: $Zn(s)\to Zn^{2 + }(aq)+2e^{-}$ is an oxidation reaction, not a cathode (reduction) reaction, so it is incorrect.
• Option C: $Ag^{+}(aq)+e^{-}\to Ag(s)$ is the reverse of the given anode reaction. For this, $E_{cell}=E^{\circ}_{Ag^{+}/Ag}-E^{\circ}_{Ag^{+}/Ag}=0$.
• Option D: For $Hg^{2+}(aq)+2e^{-}\to Hg(l)$, $E^{\circ}_{Hg^{2+}/Hg}=0.85\ V$. Then $E_{cell}=E^{\circ}_{Hg^{2+}/Hg}-E^{\circ}_{Ag^{+}/Ag}=0.85 - 0.80 = 0.05\ V>0$.

Answer:

A. $Ni^{2+}(aq)+2e^{-}\to Ni(s)$