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 determine the electromotive force (EMF) of the cell represented by the notation:

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

we use the standard electrode potentials of the two half-cells. The overall cell EMF is given by the formula:

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

Here:

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

Substituting the given values:

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

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

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

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

Answer 2

The standard EMF of the cell is calculated using: \[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] From the cell notation: - Zn is the anode $\Rightarrow E^\circ = -0.76$ V - Cu is the cathode $\Rightarrow E^\circ = +0.34$ V \[ E_{\text{cell}} = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 \ \text{V} \] Answer: $\boxed{1.10 \ \text{V}}$