Question

Given that standard electrode potential of

$Ni^{2+}$ / Ni = 0.25V, $Cu^{2+}$ / Cu = 0.334V, Ag$^+$ / Ag = 0.80V and $Zn^{2+}$ / Zn = -0.76V

Which of the following reactions under standard condition can take place in the specified direction?

• A. Ni(s) + Zn^{2+}(aq) \rightarrow Ni^{2+}(aq) + Zn(s)
• B. Cu(s) + Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + Ag(s)
• C. Cu(s) + Ni^{2+}(aq) \rightarrow Cu^{2+}(aq) + Ni(s)
• D. Ag(s) + Ni^{2+}(aq) \rightarrow Ag^{+}(aq) + Ni(s)

Answer 1

Answer: (B)

2

Answer 2

To determine which reaction can take place spontaneously under standard conditions, we calculate the standard cell potential ($E^\circ_{cell}$) for each reaction. A reaction is spontaneous if $E^\circ_{cell} > 0$.

The standard electrode potentials (reduction potentials) given are: $E^\circ_{Ni^{2+}/Ni} = 0.25V$ $E^\circ_{Cu^{2+}/Cu} = 0.334V$ $E^\circ_{Ag^{+}/Ag} = 0.80V$ $E^\circ_{Zn^{2+}/Zn} = -0.76V$

The standard cell potential is calculated as $E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation}$ or $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$. Alternatively, $E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation}$.

Let's analyze each reaction:

1. $Ni(s) + Zn^{2+}(aq) \rightarrow Ni^{2+}(aq) + Zn(s)$

   • Oxidation (Anode): $Ni(s) \rightarrow Ni^{2+}(aq) + 2e^-$ ($E^\circ_{oxidation} = -E^\circ_{Ni^{2+}/Ni} = -0.25V$)
   • Reduction (Cathode): $Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)$ ($E^\circ_{reduction} = E^\circ_{Zn^{2+}/Zn} = -0.76V$)
   • $E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} = -0.76V + (-0.25V) = -1.01V$.

   Since $E^\circ_{cell}$ is negative, this reaction is not spontaneous.

2. $Cu(s) + Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + Ag(s)$

   • Oxidation (Anode): $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$ ($E^\circ_{oxidation} = -E^\circ_{Cu^{2+}/Cu} = -0.334V$)
   • Reduction (Cathode): $Ag^{+}(aq) + e^- \rightarrow Ag(s)$ ($E^\circ_{reduction} = E^\circ_{Ag^{+}/Ag} = 0.80V$)
   • To balance electrons, the reduction half-reaction is multiplied by 2: $2Ag^{+}(aq) + 2e^- \rightarrow 2Ag(s)$.
   • The overall reaction is $Cu(s) + 2Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$.
   • $E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} = 0.80V + (-0.334V) = 0.466V$.

   Since $E^\circ_{cell}$ is positive, this reaction is spontaneous and can take place.

3. $Cu(s) + Ni^{2+}(aq) \rightarrow Cu^{2+}(aq) + Ni(s)$

   • Oxidation (Anode): $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$ ($E^\circ_{oxidation} = -E^\circ_{Cu^{2+}/Cu} = -0.334V$)
   • Reduction (Cathode): $Ni^{2+}(aq) + 2e^- \rightarrow Ni(s)$ ($E^\circ_{reduction} = E^\circ_{Ni^{2+}/Ni} = 0.25V$)
   • $E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} = 0.25V + (-0.334V) = -0.084V$.

   Since $E^\circ_{cell}$ is negative, this reaction is not spontaneous.

4. $Ag(s) + Ni^{2+}(aq) \rightarrow Ag^{+}(aq) + Ni(s)$

   • Oxidation (Anode): $Ag(s) \rightarrow Ag^{+}(aq) + e^-$ ($E^\circ_{oxidation} = -E^\circ_{Ag^{+}/Ag} = -0.80V$)
   • Reduction (Cathode): $Ni^{2+}(aq) + 2e^- \rightarrow Ni(s)$ ($E^\circ_{reduction} = E^\circ_{Ni^{2+}/Ni} = 0.25V$)
   • To balance electrons, the oxidation half-reaction is multiplied by 2: $2Ag(s) \rightarrow 2Ag^{+}(aq) + 2e^-$.
   • The overall reaction is $2Ag(s) + Ni^{2+}(aq) \rightarrow 2Ag^{+}(aq) + Ni(s)$.
   • $E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} = 0.25V + (-0.80V) = -0.55V$.

   Since $E^\circ_{cell}$ is negative, this reaction is not spontaneous.

Only reaction (2) has a positive standard cell potential, indicating it can take place spontaneously under standard conditions.