At 25°c, e° = +1.88 v for a cell based on the reaction 3 agcl(s) + al(s) → 3 ag(s) + al3+(aq) + 3 cl-(aq). find the cell potential e if [al3+] = 0.20 m and [cl-] = 0.010 m.

Use the Nernst equation which is:

E = E° - (RT/nF)*Ln Q 

Where:

E the cell voltage in V,

R the gas constant 8.314 J/K•mol,

T the room temperature normally taken 298.15K, n=3 in this case,

F the Faraday constant 96485C,

Ln is the operation of natural logarithm, and: Q = [Cl-]^3*[Al3+] 

Plugging in our values, will give us: 

E = 1.88-(8.314*298.15/(3*96485))*Ln(0.010^3...)

= 2.01 (V)

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