Question

The standard electrode potentials are: \[ E^\circ_{Cu^{2+}/Cu}=+0.34V \] \[ E^\circ_{Zn^{2+}/Zn}=-0.76V \] The standard emf of the Daniell cell is:

• A. \(0.42V\)
• B. \(0.76V\)
• C. \(1.10V\)
• D. \(1.52V\)

Hint:

Never add reduction potentials directly. Always use: \[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode} \]

Answer 1

The Correct Option is C

Concept: For a galvanic cell: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] The electrode having higher reduction potential acts as cathode.

Step 1: Identify cathode and anode. Since: \[ +0.34V > -0.76V \] Copper electrode acts as cathode. Zinc electrode acts as anode.

Step 2: Apply emf formula. \[ E^\circ_{cell} = 0.34 - (-0.76) \] \[ E^\circ_{cell} = 0.34+0.76 \] \[ E^\circ_{cell} = 1.10V \]

Step 3: Verify sign. Positive emf confirms spontaneous cell reaction. Hence: \[ E^\circ_{cell}=1.10V \] Therefore option (C) is correct.