Question

If standard reduction potential ($E^\circ$) of ($Mg^{2+} | Mg(s)$), ($Ag^+ | Ag(s)$), ($Zn^{2+}(aq) | Zn(s)$) and ($Cu^{2+}(aq) | Cu(s)$) are $-2.37$ V, $+0.79$ V, $-0.76$ V and $+0.34$ V respectively. Which of the following reaction is spontaneous?

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

Hint:

The Electrochemical Series rule of thumb: "The lower the $E^\circ$ value, the stronger the reducing agent." A metal can only displace another metal if it sits below it in the standard reduction potential table.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given standard reduction potentials for four metals and must identify which of the four proposed redox reactions will occur spontaneously.

Step 2: Key Formula or Approach:
For a redox reaction to be spontaneous, the overall standard cell potential ($E^\circ_{cell}$) must be strictly positive ($E^\circ_{cell} > 0$).
$$E^\circ_{cell} = E^\circ_{cathode \text{ (reduction)}} - E^\circ_{anode \text{ (oxidation)}}$$
Alternatively, a simpler logic rule: A metal with a more negative (lower) reduction potential will act as a reducing agent (anode) and spontaneously displace a metal with a more positive (higher) reduction potential from its aqueous solution.

Step 3: Detailed Explanation:
Let's list the potentials from most negative to most positive:
Mg ($-2.37$ V) < Zn ($-0.76$ V) < Cu ($+0.34$ V) < Ag ($+0.79$ V)
The metal on the left (more negative) can reduce the ion of any metal on its right.
Let's test the options:
- (a) Zn reducing $Mg^{2+}$: Zn ($-0.76$ V) is higher than Mg ($-2.37$ V). Zn cannot reduce Mg. (Non-spontaneous)
- (b) Ag reducing $Zn^{2+}$: Ag ($+0.79$ V) is higher than Zn. Ag cannot reduce Zn. (Non-spontaneous)
- (d) Cu reducing $Mg^{2+}$: Cu ($+0.34$ V) is higher than Mg. Cu cannot reduce Mg. (Non-spontaneous)
- (c) Zn reducing $Cu^{2+}$: Zn ($-0.76$ V) is lower/more negative than Cu ($+0.34$ V). Therefore, solid Zinc will spontaneously reduce Copper ions.
Calculation: $E^\circ_{cell} = 0.34 \text{ V (reduction)} - (-0.76 \text{ V}) \text{ (oxidation)} = +1.10 \text{ V}$. Since it is positive, it is spontaneous.

Step 4: Final Answer:
The spontaneous reaction is shown in option (c).