Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
353 views
in Chemistry by (70.4k points)
closed by
A cell is set up between copper and silver electrodes as follows:
`Cu(s)I Cu^(2+)(aq)II Ag^+(aq)Iag(S)`
If the two half cells work under standard conditions, calculate the EMF of the cell
`(Given E^(@)_(Cu^(2+)//Cu) =+0.34 V,E^(@)_(Ag^(+)//Ag)" " =+0.80 V)`

1 Answer

0 votes
by (72.2k points)
selected by
 
Best answer
From the `E^(@)` values, it is clear that copper acts as the anode and silver as the cathode.
`:.E_(cell)^(@)=E_(cathode)-E_(anode)^(@)=0.80-(0.34)=+0.46 " volt "`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...