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Anonymous asked in Science & MathematicsChemistry · 1 month ago

electrochemistry?

1)If the measured voltage of the cell Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s) is 1.37 V when the concentration of Zn2+ ion is 0.010 M, what is the Ag+ion concentration?

2)2AgBr(s) + Pb(s) → Pb2+ (aq) + 2Ag(s) + 2Br–(aq)

Each half-reaction is carried out in a separate compartment. The anion included in the

lead half-cell is NO3– The cation in the silver half-cell is K+ The two half-cells are

connected by a KNO3 salt bridge. If [Pb2+] = 1.0 M, what concentration of Br– ion will

produce a cell emf of 0.25 V at 298 K?Given AgBr(aq) + e → Ag(s) + Br-Eo= 0.0732 V

1 Answer

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  • 1 month ago
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    1) The reaction in the cell is:

    Zn(s) + 2 Ag+(aq) --> Zn2+(aq) + 2 Ag(s)

    The two half-reactions are:

    Zn(s) --> Zn2+ + 2 e- Eo = +0.76 V

    Ag+ + e- --> Ag(s) Eo = + 0.77 V

    Eo cell = 1.53 V

    Using the Nernst equation:E = Eo - RT/nF ln Q

    1.37 = 1.53 - (0.0257/2) ln (0.01 /[Ag+])

    [Ag+] = 1.7X10^-9 M

    2) The other half-cell is:Pb(s) --> Pb2+(aq) + 2 e- Eo = +0.13

    Eo cell = 0.13 + 0.073 = 0.203 V

    E = Eo - RT/nF ln Q

    Q = [Pb2+][Br-]^2

    0.25 = 0.203 - (0.0257/2) ln [1.0][Br-]^2

    [Br-] = 0.16 V

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