Discussion
Calculate the standard free energy change for the given reaction.
\(\mathrm{Ni + Cu^{2+} \to Ni^{2+} + Cu}\)
[ Given: \(E^\circ_{\mathrm{Ni^{2+}/Ni}} = -0.25\) V; \(E^\circ_{\mathrm{Cu^{2+}/Cu}} = 0.34\) V ]
Answer: C
The cell potential is the reduction potential of the cathode minus that of the anode. Copper is reduced and nickel is oxidised, so
\[ E^\circ_{cell} = 0.34 - (-0.25) = 0.59\ \text{V} \]
Two electrons are transferred (\(n = 2\)), and the Faraday constant is 96500 C/mol.
\[ \begin{aligned} \Delta G^\circ = -nFE^\circ \\ &= -2 \times 96500 \times 0.59 \\ &= -113870\ \text{J mol}^{-1} \end{aligned} \]
The sign is negative because the reaction is spontaneous. The value \(-56935\) comes from using \(n = 1\) and the positive values from the wrong sign. The answer is option C.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development