Standard cell EMF and ΔG°: Class 12 practice
E°cell = E°cathode − E°anode (reduction potentials). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Redox reactions
The idea behind standard cell emf and δg°
E°cell = E°cathode − E°anode (reduction potentials). The electrode with the higher E° is the cathode. ΔG° = −nFE°cell (F = 96500 C/mol); a positive E° means a spontaneous cell.
Step-by-step method
- Pick the higher reduction potential as the cathode.
- E°cell = E°cathode − E°anode.
- ΔG° = −nFE° (in J), divide by 1000 for kJ.
Worked example 1 (Basic)
A cell uses Mg²⁺/Mg (E° = -2.37 V) and Zn²⁺/Zn (E° = -0.76 V). Which is the cathode and what is E°cell?
- Cathode: higher E° → Zn²⁺/Zn
- E° = -0.76 − (-2.37) = 1.61 V
- Answer: Cathode Zn²⁺/Zn; 1.61 V
Worked example 2 (Standard)
A cell uses Cu²⁺/Cu (E° = 0.34 V) and Ag⁺/Ag (E° = 0.8 V). Which is the cathode and what is E°cell?
- Cathode: higher E° → Ag⁺/Ag
- E° = 0.8 − (0.34) = 0.46 V
- Answer: Cathode Ag⁺/Ag; 0.46 V
Worked example 3 (Challenge)
For the cell Mg²⁺/Mg ‖ Zn²⁺/Zn (E° = 1.61 V, n = 2), find ΔG°.
- ΔG° = −2 × 96500 × 1.61
- = -310.73 kJ/mol
- Answer: -310.73 kJ/mol
Common mistakes and misconceptions
- Adding the potentials
- Forgetting n in ΔG°
Practice questions
- (Basic) A cell uses Fe²⁺/Fe (E° = -0.44 V) and Al³⁺/Al (E° = -1.66 V). Which is the cathode and what is E°cell?
- (Basic) A cell uses Pb²⁺/Pb (E° = -0.13 V) and Ag⁺/Ag (E° = 0.8 V). Which is the cathode and what is E°cell?
- (Standard) A cell uses Mg²⁺/Mg (E° = -2.37 V) and Ni²⁺/Ni (E° = -0.25 V). Which is the cathode and what is E°cell?
- (Standard) A cell uses Cu²⁺/Cu (E° = 0.34 V) and Pb²⁺/Pb (E° = -0.13 V). Which is the cathode and what is E°cell?
- (Challenge) For the cell Ni²⁺/Ni ‖ Cu²⁺/Cu (E° = 0.59 V, n = 2), find ΔG°.
- (Challenge) For the cell Pb²⁺/Pb ‖ Ag⁺/Ag (E° = 0.93 V, n = 2), find ΔG°.
Answer key with full working
- Q1: Cathode: higher E° → Fe²⁺/Fe → E° = -0.44 − (-1.66) = 1.22 V → Answer: Cathode Fe²⁺/Fe; 1.22 V
- Q2: Cathode: higher E° → Ag⁺/Ag → E° = 0.8 − (-0.13) = 0.93 V → Answer: Cathode Ag⁺/Ag; 0.93 V
- Q3: Cathode: higher E° → Ni²⁺/Ni → E° = -0.25 − (-2.37) = 2.12 V → Answer: Cathode Ni²⁺/Ni; 2.12 V
- Q4: Cathode: higher E° → Cu²⁺/Cu → E° = 0.34 − (-0.13) = 0.47 V → Answer: Cathode Cu²⁺/Cu; 0.47 V
- Q5: ΔG° = −2 × 96500 × 0.59 → = -113.87 kJ/mol → Answer: -113.87 kJ/mol
- Q6: ΔG° = −2 × 96500 × 0.93 → = -179.49 kJ/mol → Answer: -179.49 kJ/mol
Syllabus note
This skill is listed under "Electrochemistry" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
- Electrochemistry chapter notes (Class 12)
- Cell EMF and the Nernst equation
- Faraday's laws of electrolysis
- Conductivity and molar conductivity
- Electrolysis cells in series
- Molality and mole fraction
- Boiling point elevation and freezing point depression
- First-order reactions and half-life
- Arrhenius equation
- All Class 12 practice
Frequently asked questions
Which chapter is standard cell emf and δg° in?
"Electrochemistry", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.