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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 ago
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