The table below gives the standard electrode potentials of four metal/metal-ion half-cells: Half-equation E^ / V Mg^2+(aq) + 2e^- Mg(s) -2.37 Zn^2+(aq) + 2e^- Zn(s) -0.76 Fe^2+(aq) + 2e^- Fe(s) -0.44 Cu^2+(aq) + 2e^- Cu(s) +0.34 Based only on these standard electrode potentials,…

Answer: C. Step 1: Set up the rule for feasibility For a reaction written as (metal X is oxidised) + (metal ion Y\textsuperscript{n+} is reduced) → (metal X ion) + (metal Y), the metal ion couple that is reduced is the cathode and the metal couple th…

A2 Chemistry 9701: Predicting Feasible Reactions from the Electrochemical Series (MCQ)

Half-equation	$E^{\ominus}$ / V
$\text{Mg}^{2+}(aq) + 2e^- \rightleftharpoons \text{Mg}(s)$	$-2.37$
$\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s)$	$-0.76$
$\text{Fe}^{2+}(aq) + 2e^- \rightleftharpoons \text{Fe}(s)$	$-0.44$
$\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)$	$+0.34$

Half-equation

E^{\ominus}

/ V

\text{Mg}^{2+}(aq) + 2e^- \rightleftharpoons \text{Mg}(s)

-2.37

\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s)

-0.76

\text{Fe}^{2+}(aq) + 2e^- \rightleftharpoons \text{Fe}(s)

-0.44

\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)

+0.34

Step 1: Set up the rule for feasibility

For a reaction written as (metal X is oxidised) + (metal ion Y\textsuperscript{n+} is reduced) $\rightarrow$ (metal X ion) + (metal Y), the metal ion couple that is reduced is the cathode and the metal couple that is oxidised is the anode:

$E_{cell}^{\ominus} = E^{\ominus}(\text{cathode, species reduced}) - E^{\ominus}(\text{anode, species oxidised})$

The reaction is feasible under standard conditions only if $E_{cell}^{\ominus} > 0$ .

Step 2: Test each option

A: $\text{Cu}(s) + \text{Fe}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Fe}(s)$ . Here $\text{Fe}^{2+}$ is reduced (cathode, $-0.44\ \text{V}$ ) and $\text{Cu}$ is oxidised (anode, $+0.34\ \text{V}$ ): $E_{cell}^{\ominus} = (-0.44) - (+0.34) = -0.78\ \text{V} \quad (\text{not feasible})$

B: $\text{Fe}(s) + \text{Mg}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Mg}(s)$ . Here $\text{Mg}^{2+}$ is reduced (cathode, $-2.37\ \text{V}$ ) and $\text{Fe}$ is oxidised (anode, $-0.44\ \text{V}$ ): $E_{cell}^{\ominus} = (-2.37) - (-0.44) = -1.93\ \text{V} \quad (\text{not feasible})$

C: $\text{Zn}(s) + \text{Fe}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Fe}(s)$ . Here $\text{Fe}^{2+}$ is reduced (cathode, $-0.44\ \text{V}$ ) and $\text{Zn}$ is oxidised (anode, $-0.76\ \text{V}$ ): $E_{cell}^{\ominus} = (-0.44) - (-0.76) = +0.32\ \text{V} \quad (\text{feasible, } E_{cell}^{\ominus} > 0)$

D: $\text{Cu}(s) + \text{Zn}^{2+}(aq) \rightarrow \text{Cu}^{2+}(aq) + \text{Zn}(s)$ . Here $\text{Zn}^{2+}$ is reduced (cathode, $-0.76\ \text{V}$ ) and $\text{Cu}$ is oxidised (anode, $+0.34\ \text{V}$ ): $E_{cell}^{\ominus} = (-0.76) - (+0.34) = -1.10\ \text{V} \quad (\text{not feasible})$

Step 3: Interpret the result

Only reaction C gives a positive $E_{cell}^{\ominus}$ . This matches the electrochemical series: zinc lies below iron in the table (more negative $E^{\ominus}$ ), meaning zinc is a stronger reducing agent than iron and readily loses electrons to $\text{Fe}^{2+}(aq)$ , displacing iron from solution. In every other option, the metal being oxidised is actually the weaker reducing agent of the pair (i.e. it lies above, or has a more positive $E^{\ominus}$ than, the ion it is supposedly reducing), so those reactions run in the thermodynamically unfavourable direction.

Final answer

The only feasible reaction under standard conditions is $\text{Zn}(s) + \text{Fe}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Fe}(s)$ , with $E_{cell}^{\ominus} = +0.32\ \text{V}$ , option C.