Question

The standard electrode potential of zinc electrode is \(-0.76\,V\) and that of copper electrode is \(+0.34\,V\). The standard EMF of the Daniell cell is:

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

Hint:

In galvanic cells: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Always subtract oxidation electrode potential from reduction electrode potential.

Answer 1

The Correct Option is B

Concept: The standard EMF of a galvanic cell is: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] The electrode with higher reduction potential acts as cathode.

Step 1: Identify cathode and anode.
Given: \[ E^\circ_{Zn^{2+}/Zn}=-0.76\,V \] \[ E^\circ_{Cu^{2+}/Cu}=+0.34\,V \] Since copper has higher reduction potential: \[ \text{Cathode}=Cu \] and: \[ \text{Anode}=Zn \]

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