A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10⁻⁶ M hydrogen ions. The EMF of the cell is 0.118 V at 25ºC. Calculate the concentration of hydrogen ions at the positive electrode.
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Answered by
19
The cell reaction can be written as : H2(g) + 2H+ → 2H+ + H2(g)
(I) (II) (I) (II) [ Note : (I) is negative and (II) in positive electrode ]
According to Nernst equation : Ecell = E0cell - 0.05922 log [ H+ ]2 of (I)[ H+ ]2 of (II)
Ecell = 0.0 - 0.05922 2 log 10−5x
0.118 = - 0.0592 log[ 10-5 / x ]
Then x = 10-3 M
(I) (II) (I) (II) [ Note : (I) is negative and (II) in positive electrode ]
According to Nernst equation : Ecell = E0cell - 0.05922 log [ H+ ]2 of (I)[ H+ ]2 of (II)
Ecell = 0.0 - 0.05922 2 log 10−5x
0.118 = - 0.0592 log[ 10-5 / x ]
Then x = 10-3 M
Answered by
1
Answer:
answer is 10^-4 M
Explanation:
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