Question

The standard electrode potentials of Zn and Ni are respectively $-0.76$ V and $-0.25$ V. Then the standard emf of the spontaneous cell by coupling these under standard conditions is

• A. $+1.01$ V
• B. $-0.51$ V
• C. $+0.82$ V
• D. $+0.25$ V
• E. $+0.51$ V

Hint:

Always remember: Cathode = higher reduction potential. Cell emf = Cathode − Anode (never reverse this).

Answer 1

Answer: (E)

Concept: Electrochemical cells convert chemical energy into electrical energy. The standard emf of a cell is given by: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] Where:
• Cathode = reduction occurs (gain of electrons)
• Anode = oxidation occurs (loss of electrons)
• The electrode with higher reduction potential acts as cathode

Step 1: Identify given potentials. \[ E^\circ_{Zn^{2+}/Zn} = -0.76 \text{ V} \] \[ E^\circ_{Ni^{2+}/Ni} = -0.25 \text{ V} \]

Step 2: Determine cathode and anode.
• Higher potential = more tendency to get reduced
• Ni ($-0.25$ V) > Zn ($-0.76$ V) Thus: \[ \text{Cathode} = Ni \] \[ \text{Anode} = Zn \]

Step 3: Write half-reactions. Oxidation (anode): \[ Zn \rightarrow Zn^{2+} + 2e^- \] Reduction (cathode): \[ Ni^{2+} + 2e^- \rightarrow Ni \]

Step 4: Calculate emf. \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ E^\circ = (-0.25) - (-0.76) \] \[ E^\circ = -0.25 + 0.76 = 0.51 \text{ V} \]

Step 5: Check spontaneity.
• Positive emf → spontaneous reaction
• Hence reaction will proceed in given direction Final Answer: \[ \boxed{0.51 \text{ V}} \]