Question

In an electrochemical cell, the standard electrode potential of \( \text{Zn}^{2+}/\text{Zn} \) is \( -0.76 \, \text{V} \) and that of \( \text{Cu}^{2+}/\text{Cu} \) is \( +0.34 \, \text{V} \). What is the standard EMF of the cell formed by these electrodes?

• A. \( 0.42 \, \text{V} \)
• B. \( 1.10 \, \text{V} \)
• C. \( -1.10 \, \text{V} \)
• D. \( -0.42 \, \text{V} \)

Hint:

To calculate the EMF of a galvanic cell, subtract the standard reduction potential of the anode from that of the cathode. A positive EMF indicates a spontaneous cell reaction.

Answer 1

The Correct Option is B

The electromotive force (EMF) of an electrochemical cell is calculated using the standard electrode potentials of the cathode and anode: \[ \text{EMF} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Where: - \( E^\circ_{\text{cathode}} \) is the standard reduction potential of the electrode where reduction occurs, - \( E^\circ_{\text{anode}} \) is the standard reduction potential of the electrode where oxidation occurs. Given: - Standard electrode potential of \( \text{Zn}^{2+}/\text{Zn} = -0.76 \, \text{V} \), - Standard electrode potential of \( \text{Cu}^{2+}/\text{Cu} = +0.34 \, \text{V} \). In a galvanic cell, the electrode with the higher reduction potential undergoes reduction (acts as the cathode), and the electrode with the lower reduction potential undergoes oxidation (acts as the anode). Comparing the potentials: - \( E^\circ (\text{Cu}^{2+}/\text{Cu}) = +0.34 \, \text{V} \) (higher, so cathode), - \( E^\circ (\text{Zn}^{2+}/\text{Zn}) = -0.76 \, \text{V} \) (lower, so anode). The cell can be represented as: \[ \text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu} \] Calculate the EMF: \[ \text{EMF} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \] \[ \text{EMF} = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 \, \text{V} \] The positive EMF indicates that the cell is spontaneous as written. Thus, the standard EMF of the cell is \( 1.10 \, \text{V} \).