Question

The standard electrode potential for $\text{Zn}^{2+}/\text{Zn}$ is $-0.76$ V and for $\text{Cu}^{2+}/\text{Cu}$ is $+0.34$ V. The EMF of the cell
$$ \text{Zn}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu} $$ is:

• A. 1.10 V
• B. 0.76 V
• C. -0.42 V
• D. 0.34 V

Hint:

EMF of a galvanic cell is always cathode potential minus anode potential. The more positive E° value is usually the cathode.

Answer 1

The Correct Option is A

To calculate the electromotive force (EMF) of the electrochemical cell denoted as:

\[\text{Zn}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu}\]

The standard electrode potentials of the constituent half-cells are utilized. The overall cell EMF is determined using the equation:

\[E_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]

Where:

• \(E^{\circ}_{\text{cathode}} = +0.34 \, \text{V}\) (corresponding to the \(\text{Cu}^{2+}/\text{Cu}\) half-reaction)
• \(E^{\circ}_{\text{anode}} = -0.76 \, \text{V}\) (corresponding to the \(\text{Zn}^{2+}/\text{Zn}\) half-reaction)

Upon substitution of these values:

\[E_{\text{cell}} = +0.34 \, \text{V} - (-0.76 \, \text{V})\]

\[E_{\text{cell}} = +0.34 \, \text{V} + 0.76 \, \text{V}\]

\[E_{\text{cell}} = 1.10 \, \text{V}\]

Consequently, the EMF of this cell is \(1.10 \, \text{V}\).