JEE Main 2025 · ChemistryNumerical answerNumerical valueHardMulti-step
JEE Main 7 April 2025, Shift 1, Chemistry Q71
Question 71 of 75 in this shift, Chemistry question 21 of 25, Section B.
1 Faraday electricity was passed through (1.5 M, 1 L) /Cu and 0.1 Faraday was passed through (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is __________ mV (nearest integer)
Given : V
V
V
The figure, in words
Electrochemical cell: Cu electrode dipping in Cu2+(aq) beaker (left) and Ag electrode dipping in Ag+(aq) beaker (right), joined by a salt bridge; the two electrodes are connected through a galvanometer/meter at the top.Official answer
400
NTA final key.
- Chapter
- Electrochemistry
- Topic
- Nernst Equation
- Idea tested
- Full Cell Nernst Equation