Question

Consider an electrochemical cell involving the overall reaction

2 AgBr(s) + Pb(s) → Pb2+(aq) + 2 Ag(s) + 2 Br–(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 and [Br-] = 0.14 M, what is the emf of the cell at 298 K?
Given:

AgBr(s) + e– → Ag(s) + Br–(aq)              E° = +0.07 V.

Pb⁺² + 2e– → Pb(s)                                E° = - 0.13 V.

Answer 1

Step 1: Find 'E'

Reduction semi-cell:

AgBr(s) + e– → Ag(s) + Br–(aq)           E° = + 0.07 V

Oxidation semi-cell:

Pb(s) --> Pb⁺² + 2e–    E = + 0.13 V (note that I turned the sign because I turned the equation too)

To balance the cell reaction multiply reduction semi-cell by two:

(AgBr(s) + e– → Ag(s) + Br–(aq)) x2

Pb(s) --> Pb⁺² + 2e–

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

and E = 0.07 + 0.13 = 0.20 V

(Some people prefer the method E = Ered - Eoxi. In this case you don't have to turn the signs but just recognize which semi-cell is doing what. Note we can get the same result --> E = 0.07 - (-0.13) = 0.20)

Step 2: Apply Nernst equation

Where,

R = 8.314 J/mol K

n = moles of electrons (2 in this case)

F = Faraday's constant, or 96500 C/mol

Q = [products]^p/[reagents]^r (only ionic compounds/elements)

so, Q = [Br-]² * [Pb+2] = 0.14² * 1.0 = 0.0196

T = 298 K

When you're using this kind of equations put special care in the units. Any variation will result in wrong results. If you're using SI keep all units in that system.

Replacing,

E = 0.20 V - ((8.314J/mol K * 298 K)/(2mol *96500C/mol))*ln(0.0196) = 0.25 V