Question

What is the standard emf of the following cell?
\[ \mathrm{Ni(s)\ | \ Ni^{2+}(aq)\ || \ Au^{3+}(aq)\ | \ Au(s)} \] \[ E^\circ_{\mathrm{Ni}} = -0.25\ \mathrm{V}, \quad E^\circ_{\mathrm{Au}} = +1.50\ \mathrm{V} \]

• A. –1.25 V
• B. 1.75 V
• C. 1.25 V
• D. –1.75 V

Hint:

Cell emf is always calculated as cathode potential minus anode potential.

Answer 1

The Correct Option is B

Step 1: Identifying cathode and anode.
Gold has higher reduction potential, so it acts as cathode. Nickel acts as anode.

Step 2: Using emf formula.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Step 3: Substituting values.
\[ E^\circ_{\text{cell}} = 1.50 - (-0.25) = 1.75\ \mathrm{V} \]

Step 4: Conclusion.
The standard emf of the cell is 1.75 V.