when two half-cells of electrode potential of `E_1` and `E_2` are combined to form a half cell of electrode potential `E_3` , then (when `n_1,n_2"and"n_3` are no. of electrons exchanged in first second and combined half-cells:
A
`E_3=E_2-E_1`
B
`E_3=(E_1n_1+E_2n_2)/(n_3)`
C
`E_3=(E_1n_1+E_2n_2)/(n_3^2)`
D
`E_3=E_1-E_2`
Text Solution
Verified by Experts
The correct Answer is:
B
Topper's Solved these Questions
ELECTROCHEMISTRY
NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|1 Videos
ELECTROCHEMISTRY
NARENDRA AWASTHI|Exercise Level 1 (Q.91 To Q.120)|1 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
For the half cell At pH = 3 electrode potential is
For the half cell At pH=2 . Electrod potential is :
For the half cell At pH=2 , the electrode potential is
The half-cell reduction potential of a hudrogen electrode at pH= 10 will be.
The half cell reduction potential of a hydrogen electrode at pH = 5 will be :
Prove that for two half reactions having potentials E_(1) and E_(2) which are combined to yield a third half reaction, having a potential E_(3) , E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)
The half cell reduction potential of a hydrogen electrode at pH= 10 and H_2 gas at 1 atmi will be
