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)