Question

question 23 of 30
the half - reaction at the cathode in an electrochemical cell is given below. what other half - reaction would most likely occur at the anode to produce a spontaneous reaction?
ni^{2 + }(aq)+2e^{-}
ightarrow ni(s)
click for a reduction potential chart
a. cu(s)
ightarrow cu^{2 + }(aq)+2e^{-}
b. mg^{2 + }(aq)+2e^{-}
ightarrow mg(s)
c. zn(s)
ightarrow zn^{2 + }(aq)+2e^{-}
d. ni(s)
ightarrow ni^{2 + }(aq)+2e^{-}

Explanation:

Step1: Recall anode - cathode reactions

Oxidation occurs at the anode and reduction at the cathode. The given cathode reaction is $Ni^{2 + }(aq)+2e^{-}\to Ni(s)$ (a reduction). We need an oxidation reaction at the anode with a suitable standard - reduction potential to give a positive $E_{cell}^{\circ}$ for a spontaneous reaction.

Step2: Consider standard - reduction potentials

The standard - reduction potential of $Ni^{2+}/Ni$ is $E^{\circ}_{Ni^{2+}/Ni}=- 0.25$ V. For a spontaneous reaction, $E_{cell}^{\circ}=E^{\circ}_{cathode}-E^{\circ}_{anode}>0$. We need an oxidation reaction (reverse of a reduction half - reaction) with a more negative $E^{\circ}$ value than that of $Ni^{2+}/Ni$.
The standard - reduction potential of $Cu^{2+}/Cu$ is $E^{\circ}_{Cu^{2+}/Cu}=0.34$ V, for $Mg^{2+}/Mg$ is $E^{\circ}_{Mg^{2+}/Mg}=-2.37$ V, for $Zn^{2+}/Zn$ is $E^{\circ}_{Zn^{2+}/Zn}=-0.76$ V.

Step3: Analyze each option

• Option A: The reverse of the $Cu^{2+}/Cu$ reduction has $E^{\circ}_{Cu^{2+}/Cu}=0.34$ V. If $Cu(s)\to Cu^{2+}(aq) + 2e^{-}$ occurs at the anode, $E_{cell}^{\circ}=-0.25 - 0.34=-0.59$ V (non - spontaneous).
• Option B: $Mg^{2+}(aq)+2e^{-}\to Mg(s)$ is a reduction, not an oxidation (rejected as it's not an anode reaction).
• Option C: The reverse of the $Zn^{2+}/Zn$ reduction is $Zn(s)\to Zn^{2+}(aq)+2e^{-}$ with $E^{\circ}_{Zn^{2+}/Zn}=-0.76$ V. $E_{cell}^{\circ}=-0.25-(-0.76)=0.51$ V (spontaneous).
• Option D: This is the reverse of the given cathode reaction and will not give a spontaneous reaction as it would result in $E_{cell}^{\circ}=0$.

Answer:

C. $Zn(s)\to Zn^{2+}(aq)+2e^{-}$