Question

Calculate the standard potential of the cell that has the following reaction.
\(\text{Zn}_{(s)}+2\text{H}_{(1M)}^{+}\rightarrow \text{H}_{2(g,1\text{atm})}+\text{Zn}_{(1M)}^{2+}\) if \(E^{0}(\text{Zn}^{2+}/\text{Zn})=-0.763\) V.

• A. \(0.763\) V
• B. \(0.381\) V
• C. \(-0.763\) V
• D. \(0.14\) V

Hint:

Zinc is oxidised at the anode and hydrogen ion is reduced at the standard hydrogen electrode, so E cell is E cathode minus E anode.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The cell reaction is \(\text{Zn}+2\text{H}^+\rightarrow\text{H}_2+\text{Zn}^{2+}\). Zinc is oxidised (anode) and \(\text{H}^+\) is reduced (cathode).

Step 2: Electrode potentials
\(E^0_{\text{H}^+/\text{H}_2}=0\) V for the standard hydrogen electrode, and \(E^0_{\text{Zn}^{2+}/\text{Zn}}=-0.763\) V.

Step 3: Calculation
\(E^0_{cell}=E^0_{cathode}-E^0_{anode}=0-(-0.763)=0.763\) V.

Step 4: Compare the options
This is option (A).

Final Answer:
The standard potential of the cell is 0.763 V. \[ \boxed{\text{(A) }0.763\text{ V}} \]