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 25° C. Calculate the concentration of hydrogen ions at the positive electrode.

Question

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 25° C. Calculate the concentration of hydrogen ions at the positive electrode. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∵ Emf =0.118 V >0, it is galvanic cell and anode is negative electrode: At anode: H 2(g) longrightarrow 2 H +(10-6 M )+2 e- At cathode: 2 H +(x)+2 e- longrightarrow H 2 Cell reaction: H +(x) longrightarrow H +(10-6 M ) Emf =0.118 V =0-0.0592 log (10-6/x) ⇒ x=10-4 M

Q. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10−6M hydrogen ions. The emf of the cell is 0.118V at 25∘C. Calculate the concentration of hydrogen ions at the positive electrode.

∵ Emf =0.118V>0, it is galvanic cell and anode is negative electrode : At anode : H2​(g)⟶2H+(10−6M)+2e− At cathode :2H+(x)+2e−⟶H2​ Cell reaction : H+(x)⟶H+(10−6M) Emf =0.118V=0−0.0592logx10−6​ ⇒x=10−4M

Q. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $1 0^{- 6} M$ hydrogen ions. The emf of the cell is $0.118 V$ at $2 5^{\circ} C .$ Calculate the concentration of hydrogen ions at the positive electrode.