JEE Main Chemistry question
The emf of the cell \mathrm{Ni\,|\,Ni^{2+}(1.0\ M)\,||\,Au^{3+}(1.0\ M)\,|\,Au} is [ V and V]
Answer: (D)
Solution
Ni (lower reduction potential) is the anode and Au the cathode.
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account