Question

The spontaneous reaction that takes place in this cell is

• A. \(\mathrm{Zn} + \mathrm{Ni} \rightarrow \mathrm{Zn^{2+}} + \mathrm{Ni^{2+}}\)
• B. \(\mathrm{Zn} + \mathrm{Ni^{2+}} \rightarrow \mathrm{Zn^{2+}} + \mathrm{Ni}\)
• C. \(\mathrm{Ni} + \mathrm{Zn^{2+}} \rightarrow \mathrm{Ni^{2+}} + \mathrm{Zn}\)
• D. \(\mathrm{Ni^{2+}} + 2\mathrm{K} \rightarrow 2\mathrm{K^+} + \mathrm{Ni}\)

Hint:

Metal with lower reduction potential acts as anode and undergoes oxidation.

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}>0 \]
Step 2: Calculation / Simplification}
\(E^\circ_{\mathrm{Zn^{2+}/Zn}} = -0.76\ \mathrm{V}\)
\(E^\circ_{\mathrm{Ni^{2+}/Ni}} = -0.25\ \mathrm{V}\)
\(\mathrm{Zn}\) is more easily oxidized (lower reduction potential).
Spontaneous: \(\mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^-\) (anode)
\(\mathrm{Ni^{2+}} + 2e^- \rightarrow \mathrm{Ni}\) (cathode)
Overall: \(\mathrm{Zn} + \mathrm{Ni^{2+}} \rightarrow \mathrm{Zn^{2+}} + \mathrm{Ni}\)
Step 3: Final Answer
\[ \mathrm{Zn} + \mathrm{Ni^{2+}} \rightarrow \mathrm{Zn^{2+}} + \mathrm{Ni} \]