Question
In: Chemistry
A galvanic cell consists of a lead electrode in 0.0256 M Pb(NO3)2 and a aluminum electrode...
A galvanic cell consists of a lead electrode in 0.0256 M Pb(NO3)2 and a aluminum electrode in 0.292 M Al(NO3)3. What is the cell potential (emf, in V) of this cell at 25oC?
Solutions
Expert Solution
2Al(s) -------------------> 2Al^3+(aq) + 6e^- E0 = 1.66v
3Pb^2+ (aq) + 6e^- -----------> 3Pb(s) E0 = -0.13v
----------------------------------------------------------------------------------------
2Al(s) + 3Pb^2+ (aq) ---------> 2Al^3+ (aq) + 3Pb(s) E0cell = 1.53v
n = 6
Ecell = E0cell - 0.0592/n logQ
= 1.53 - 0.0592/6 log[Al^3+]^2/[Pb^2+]^3
= 1.53 - 0.00986log(0.292)^2/(0.0256)^3
= 1.53 - 0.00986 *3.7060
= 1.4934V >>>>>>answer
queen_honey_blossom answered 2 years agoRelated Solutions
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